Problems and Solutions Projectile Motion
Soal #1
A rifle is aimed horizontally at a target 30 m away. The bullet hits the target 1.9 cm below the aiming point. (a) What is the bullet’s time of flight? (b) What is the muzzle velocity?Answer:
(a) First, we define our coordinates. I will use the coordinate system indicated in Fig. 3.1, where the origin is placed at the tip of the gun. Then we have x0 = 0 and y0 = 0. We also know the acceleration:
ax = 0 and ay = −9.80 m/s2 = − g
What else do we know? The gun is fired horizontally so that v0y = 0, but we don’t know v0x. We don’t know the time of flight but we do know that when x has the value 30 m then y has the value −1.9×10−2 m. (Minus!)
Our equation for the y coordinate is
y = y0 + v0yt – ½gt2
(−1.9×10−2 m) = 0 + 0.t – 0.5(9.8 m/s2)t2
t2 = 2(−1.9×10−2 m)/(9.8 m/s2)
t2 = 3.9 x 10-3 s2
t = 6.2 x 10-2 s
Since this is the time of impact with the target, the time of flight of the bullet is t = 6.2×10−2 s.
(b) The equation for x−motion is
x = v0xt
v0x = x/t = 30 m/(6.2 x 10-2 s)
v0x = 480 m/s
The muzzle velocity of the bullet is 480 m/s.
Problem #2
In a local bar, a customer slides an empty beer mug on the counter for a refill. The bartender does not see the mug, which slides off the counter and strikes the floor 1.40m from the base of the counter. If the height of the counter is 0.860 m, (a) with what speed did the mug leave the counter and (b) what was the direction of the mug’s velocity just before it hit the floor?
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| Fig.1 |
Answer:
(a) The motion of the beer mug is shown in Fig.1(a). We choose the origin of our xy coordinate system as being at the point where the mug leaves the counter. So the mug’s initial coordinates for its flight are x0 = 0, y0 = 0.
At the very beginning of its motion through the air, the velocity of the mug is horizontal. (This is because its velocity was horizontal all the time it was sliding on the counter.) So we know that v0y = 0 but we don’t know the value of v0x. (In fact, that’s what we’re trying to figure out!)
We might begin by finding the time t at which the mug hit the floor. This is the time t at which y = −0.860m (recall how we chose the coordinates!), and we will need the y equation of motion for this; since v0y = 0 and ay = −g, we get:
−0.860 m = 0 + 0.t – ½(9.8 m/s2)t2
t2 = 2(0.860)/(9.8) = 0.176 s2
t = 0.419 s
is the time of impact. To find v0x we consider the x equation of motion; x0 = 0 and ax = 0, so we have
x = v0xt
At t = 0 .419 s we know that the x coordinate was equal to 1.40 m. So
1.40 m = v0x (0.419 s)
Solve for v0x:
v0x = 1.40/0.419 = 3.34 m/s
which tells us that the initial speed of the mug was v0 = 3.34 m/s.
(b) We want to find the components of the mug’s velocity at the time of impact, that is, at t = 0 .419 . Substitute into our expressions for vx and vy:
vx = v0x = 3.34 m/s
and
vy = v0y – gt = 0 – (9.8 m/s2)(0.419 s) = –4.11 m/s
So at the time of impact, the speed of the mug was
v = (vx2 + vy2)1/2 = [(3.34 m/s)2 + (–4.11 m/s)2]1/2 = 5.29 m/s
and, if as in Fig. 1(b) we let θ be the angle below the horizontal at which the velocity vector is pointing, we see that
tan θ = 4.11/3.34 = 1.23
θ = tan-1(1.23) = 50.90
At the time of impact, the velocity of the mug was directed at 50.90 below the horizontal.
Problem #3
You throw a ball with a speed of 25.0 m s at an angle of 40.00 above the horizontal directly toward a wall, as shown in Fig. 2. The wall is 22.0m from the release point of the ball. (a) How long does the ball take to reach the wall? (b) How far above the release point does the ball hit the wall? (c) What are the horizontal and vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?
(a) The motion of the beer mug is shown in Fig.1(a). We choose the origin of our xy coordinate system as being at the point where the mug leaves the counter. So the mug’s initial coordinates for its flight are x0 = 0, y0 = 0.
At the very beginning of its motion through the air, the velocity of the mug is horizontal. (This is because its velocity was horizontal all the time it was sliding on the counter.) So we know that v0y = 0 but we don’t know the value of v0x. (In fact, that’s what we’re trying to figure out!)
We might begin by finding the time t at which the mug hit the floor. This is the time t at which y = −0.860m (recall how we chose the coordinates!), and we will need the y equation of motion for this; since v0y = 0 and ay = −g, we get:
−0.860 m = 0 + 0.t – ½(9.8 m/s2)t2
t2 = 2(0.860)/(9.8) = 0.176 s2
t = 0.419 s
is the time of impact. To find v0x we consider the x equation of motion; x0 = 0 and ax = 0, so we have
x = v0xt
At t = 0 .419 s we know that the x coordinate was equal to 1.40 m. So
1.40 m = v0x (0.419 s)
Solve for v0x:
v0x = 1.40/0.419 = 3.34 m/s
which tells us that the initial speed of the mug was v0 = 3.34 m/s.
(b) We want to find the components of the mug’s velocity at the time of impact, that is, at t = 0 .419 . Substitute into our expressions for vx and vy:
vx = v0x = 3.34 m/s
and
vy = v0y – gt = 0 – (9.8 m/s2)(0.419 s) = –4.11 m/s
So at the time of impact, the speed of the mug was
v = (vx2 + vy2)1/2 = [(3.34 m/s)2 + (–4.11 m/s)2]1/2 = 5.29 m/s
and, if as in Fig. 1(b) we let θ be the angle below the horizontal at which the velocity vector is pointing, we see that
tan θ = 4.11/3.34 = 1.23
θ = tan-1(1.23) = 50.90
At the time of impact, the velocity of the mug was directed at 50.90 below the horizontal.
Problem #3
You throw a ball with a speed of 25.0 m s at an angle of 40.00 above the horizontal directly toward a wall, as shown in Fig. 2. The wall is 22.0m from the release point of the ball. (a) How long does the ball take to reach the wall? (b) How far above the release point does the ball hit the wall? (c) What are the horizontal and vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?
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| Fig.2 |
Answer:
(a) We will use a coordinate system which has its origin at the point of firing, which we take to be at ground level. What is the mathematical condition which determines when the ball hits the wall? It is when the x coordinate of the ball is equal to 22.0 m. Then let’s write out the x−equation of motion for the ball. The ball’s initial x− velocity is
v0x = v0 cos θ0 = (25.0 m/s) cos 40.00 = 19.2 m/s
and of course ax = 0, so that the x motion is given by
x = v0xt =
22.0 m = (19.2 m/s)t
t = 22.0/19.2 = 1.15 s
The ball hits the wall 1.15s after being thrown.
(b) We will be able to answer this question if we can find the y coordinate of the ball at the time that it hits the wall, namely at t = 1.15 s. We need the y equation of motion. The initial y velocity of the ball is
v0y = v0 sin θ0 = (25.0 m/s) cos 40.00 = 16.1 m/s
and the y acceleration of the ball is ay =−g giving:
y = y0 + v0yt – ½gt2
= 0 + (16.1 m/s)(1.15 s) – 0.5(9.8 m/s2)(1.15 s)2
y = 12.0 m
which tells us that the ball hits the wall at 12.0 m above the ground level (above the release point).
(c) The x and y components of the balls’s velocity at the time of impact, namely at t = 1.15s are found from Eqs.
vx = v0x = 19.2 m/s
and
vy = v0y – gt = 16.1 m/s + (−9.80 m/s2)(1.15 s) = +4.83 m/s
(d) Has the ball already passed the highest point on its trajectory? Suppose the ball was on its way downward when it struck the wall. Then the y component of the velocity would be negative, since it is always decreasing and at the trajectory’s highest point it is zero. (Of course, the x component of the velocity stays the same while the ball is in flight.) Here we see that the y component of the ball’s velocity is still positive at the time of impact. So the ball was still climbing when it hit the wall; it had not reached the highest point of its (free) trajectory.
Problem #4
The launching speed of a certain projectile is five times the speed it has at its maximum height. Calculate the elevation angle at launching.
(a) We will use a coordinate system which has its origin at the point of firing, which we take to be at ground level. What is the mathematical condition which determines when the ball hits the wall? It is when the x coordinate of the ball is equal to 22.0 m. Then let’s write out the x−equation of motion for the ball. The ball’s initial x− velocity is
v0x = v0 cos θ0 = (25.0 m/s) cos 40.00 = 19.2 m/s
and of course ax = 0, so that the x motion is given by
x = v0xt =
22.0 m = (19.2 m/s)t
t = 22.0/19.2 = 1.15 s
The ball hits the wall 1.15s after being thrown.
(b) We will be able to answer this question if we can find the y coordinate of the ball at the time that it hits the wall, namely at t = 1.15 s. We need the y equation of motion. The initial y velocity of the ball is
v0y = v0 sin θ0 = (25.0 m/s) cos 40.00 = 16.1 m/s
and the y acceleration of the ball is ay =−g giving:
y = y0 + v0yt – ½gt2
= 0 + (16.1 m/s)(1.15 s) – 0.5(9.8 m/s2)(1.15 s)2
y = 12.0 m
which tells us that the ball hits the wall at 12.0 m above the ground level (above the release point).
(c) The x and y components of the balls’s velocity at the time of impact, namely at t = 1.15s are found from Eqs.
vx = v0x = 19.2 m/s
and
vy = v0y – gt = 16.1 m/s + (−9.80 m/s2)(1.15 s) = +4.83 m/s
(d) Has the ball already passed the highest point on its trajectory? Suppose the ball was on its way downward when it struck the wall. Then the y component of the velocity would be negative, since it is always decreasing and at the trajectory’s highest point it is zero. (Of course, the x component of the velocity stays the same while the ball is in flight.) Here we see that the y component of the ball’s velocity is still positive at the time of impact. So the ball was still climbing when it hit the wall; it had not reached the highest point of its (free) trajectory.
Problem #4
The launching speed of a certain projectile is five times the speed it has at its maximum height. Calculate the elevation angle at launching.
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| Fig.3 |
Answer:
We make a diagram of the projectile’s motion in Fig. 3. The launch it speed is v0, and the projectile is launched at an angle θ0 upward from the horizontal. We might start this problem by solving for the time it takes the projectile to get to maximum height, but we can note that at maximum height, there is no y velocitycomponent, and the x velocity component is the same as it was when the projectile was launched. Therefore at maximum height the velocity components are
vx = v0 cos θ0 and vy = 0
and so the speed of the projectile at maximum height is v0 cos θ0. Now, we are told that the launching speed (v0) is five times the speed at maximum height. This gives us:
v0 =5 v0 cosθ0 ⇒ cos θ0 = 1/5
which has the solution
θ0 = 78.50
So the elevation angle at launching is θ0 = 78.50
Problem #5
During volcanic eruptions, chunks of solid rock can be blasted out of a volcano; these projectiles are called volcanic bombs. Fig. 4, shows a cross section of Mt. Fuji in Japan. (a) At what initial speed would the bomb have to be ejected, at 350 to the horizontal, from the vent at A in order to fall at the foot of the volcano at B? (Ignore the effects of air on the bomb’s travel.) (b) What would be the time of flight?
We make a diagram of the projectile’s motion in Fig. 3. The launch it speed is v0, and the projectile is launched at an angle θ0 upward from the horizontal. We might start this problem by solving for the time it takes the projectile to get to maximum height, but we can note that at maximum height, there is no y velocitycomponent, and the x velocity component is the same as it was when the projectile was launched. Therefore at maximum height the velocity components are
vx = v0 cos θ0 and vy = 0
and so the speed of the projectile at maximum height is v0 cos θ0. Now, we are told that the launching speed (v0) is five times the speed at maximum height. This gives us:
v0 =5 v0 cosθ0 ⇒ cos θ0 = 1/5
which has the solution
θ0 = 78.50
So the elevation angle at launching is θ0 = 78.50
Problem #5
During volcanic eruptions, chunks of solid rock can be blasted out of a volcano; these projectiles are called volcanic bombs. Fig. 4, shows a cross section of Mt. Fuji in Japan. (a) At what initial speed would the bomb have to be ejected, at 350 to the horizontal, from the vent at A in order to fall at the foot of the volcano at B? (Ignore the effects of air on the bomb’s travel.) (b) What would be the time of flight?
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| Fig.4 |
Answer:
(a) We use a coordinate system with its origin at point A (the volcano “vent”); then for the flight from the vent at A to point B, the initial coordinates are x0 = 0 and y0 = 0, and the final coordinates are x = 9.40km and y = −3.30 km. Aside from this, we don’t know the initial speed of the rock (that’s what we’re trying to find) or the time of flight from A to B. Of course, the acceleration of the rock is given by ax = 0, ay = −g. We start with the x equation of motion. The initial x−velocity is
v0x = v0 cos θ
where θ = 350 so the functions x(t) is
x = v0xt = (v0 cos θ)t
Now, we do know that at the time of impact x had the value x = 9 .40 km so if we now let t be the time of flight, then
(9.40km) = v0 cosθ t or t = (9.40km)/(v0 cos θ) (*)
Next we look at the y equation of motion. Since v0y = v0 sin θ we get:
y = y0 + v0yt – ½ gt2
y = 0 + v0 sin θt – ½gt2
But at the time t of impact the y coordinate had the value y = −3.30 km. If we also substitute for t in this expression using Eq. (*) we get:
−3.30 km = v0 sin θ[9.40 km/v0 cos θ] – ½ g[9.40 km/v0 cosθ]2
= (9 .40 km)tanθ – [g(9.40km)2/(2v02 cos2θ)]
At this point we are done with the physics problem. The only unknown in this equation is v0, which we can find by doing a little algebra:
[g(9.40km)2/(2v02 cos2θ)] = (9 .40 km) tanθ +3 .30 km
which gives:
v02 = g(9.40km)2/9.88 cos2θ)
v02 = (9.8 m/s2)(0.951 km)/(cos2 350)
= 1.39 x 104 m2/s2
v0 = 118 m/s
(b) Having v0 in hand, finding t is easy. Using our result from part(a) and Eq. (*) we find:
t = (9.40km)/v0 cosθ
= (9400m)/(118 m/s x cos 350)
t = 97.2 s
The time of flight is 97.2s.
Problem #5
A horizontal rifle is fired at a bull’s–eye. The muzzle speed of the bullet is 670 m/s . The barrel is pointed directly at the center of the bull’s–eye, but the bullet strikes the target 0.025 m below the center. What is the horizontal distance between the end of the rifle and the bull’s–eye?
v0x = v0 cos θ
where θ = 350 so the functions x(t) is
x = v0xt = (v0 cos θ)t
Now, we do know that at the time of impact x had the value x = 9 .40 km so if we now let t be the time of flight, then
(9.40km) = v0 cosθ t or t = (9.40km)/(v0 cos θ) (*)
Next we look at the y equation of motion. Since v0y = v0 sin θ we get:
y = y0 + v0yt – ½ gt2
y = 0 + v0 sin θt – ½gt2
But at the time t of impact the y coordinate had the value y = −3.30 km. If we also substitute for t in this expression using Eq. (*) we get:
−3.30 km = v0 sin θ[9.40 km/v0 cos θ] – ½ g[9.40 km/v0 cosθ]2
= (9 .40 km)tanθ – [g(9.40km)2/(2v02 cos2θ)]
At this point we are done with the physics problem. The only unknown in this equation is v0, which we can find by doing a little algebra:
[g(9.40km)2/(2v02 cos2θ)] = (9 .40 km) tanθ +3 .30 km
which gives:
v02 = g(9.40km)2/9.88 cos2θ)
v02 = (9.8 m/s2)(0.951 km)/(cos2 350)
= 1.39 x 104 m2/s2
v0 = 118 m/s
(b) Having v0 in hand, finding t is easy. Using our result from part(a) and Eq. (*) we find:
t = (9.40km)/v0 cosθ
= (9400m)/(118 m/s x cos 350)
t = 97.2 s
The time of flight is 97.2s.
Problem #5
A horizontal rifle is fired at a bull’s–eye. The muzzle speed of the bullet is 670 m/s . The barrel is pointed directly at the center of the bull’s–eye, but the bullet strikes the target 0.025 m below the center. What is the horizontal distance between the end of the rifle and the bull’s–eye?
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| Fig.5 |
As usual, begin by drawing a picture of what is happening! The problem is diagrammed in Figure. Even though the rifle is pointed straight at the bull’s–eye, the bullet must miss because it will take a certain amount of time to travel the horizontal distance to the target and in that time the bullet will have some downward vertical motion. The rifle was fired horizontally, and from that we know:
v0x = 670 m/s; v0y = 0
We know that at the time the bullet struck the target its y coordinate was y = −0.025 m. Then using the y part of Eq,
y = y0 + v0yt – ½ gt2
we get:
−0.025 m = 0 - ½ (9.8 m/s2)t2 = 0
which gives us
t2 = 2(−0.025 m)/(−9.80 m/s2) = 5.1 × 10−3 s2
t = 7.1 x 10-3 s
The distance to the target is the value of x at the time the bullet struck. Now that we have the time of impact we find x using the x part of Eq,
x = v0xt = (670 m/s)( 7.1 x 10-3 s) = 48 m
Problem #6
A golf ball rolls off a horizontal cliff with an initial speed of 11.4 m/s. The ball falls a vertical distance of 15.5 m into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water?
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| Fig.6 |
Answer:
(a) We draw picture of the ball and its motion along with the coordinates, as in Fig. 6. Since the y axis goes upward, the level of the water is at y = −15.5 m. Note that the ball rolls off the cliff horizontally, so that it has an initial x velocity: v0x = 11.4 m/s , but there is no initial y velocity: v0y = 0. To answer (a) we think about the mathematical condition that the ball has hit the water. This is when y = −15.5 m. (We don’t know the x coordinate of the ball when it hits the water.) Then using the y part of Eq. (*) with ay = −g and v0y we can solve for the time t:
y = y0 + v0yt + ½ayt2
−15.5 m = 0 + 1/2 (−9.80 m/s2)t2
t2 = 2(−15.5 m)/(−9.80 m/s2) = 3.2 s2
t = 1.8 s
(b) If we have both components of the velocity at the time the ball hits the water, we can find the speed from Eq,
vx = v0x = 11.4 m/s and
vy = v0y + ayt
= 0 + (–9.8 m/s2)(1.8 s)
vy = –17.4 m/s
Then the speed of the ball at impact is
v = [vx2 + vy2]1/2
v = [(11.4 m/s)2 + (−17.4 m/s)2]1/2 = 20.8 m/s
The speed of the ball when it hits the water is 20.8 m/s .
(a) We draw picture of the ball and its motion along with the coordinates, as in Fig. 6. Since the y axis goes upward, the level of the water is at y = −15.5 m. Note that the ball rolls off the cliff horizontally, so that it has an initial x velocity: v0x = 11.4 m/s , but there is no initial y velocity: v0y = 0. To answer (a) we think about the mathematical condition that the ball has hit the water. This is when y = −15.5 m. (We don’t know the x coordinate of the ball when it hits the water.) Then using the y part of Eq. (*) with ay = −g and v0y we can solve for the time t:
y = y0 + v0yt + ½ayt2
−15.5 m = 0 + 1/2 (−9.80 m/s2)t2
t2 = 2(−15.5 m)/(−9.80 m/s2) = 3.2 s2
t = 1.8 s
(b) If we have both components of the velocity at the time the ball hits the water, we can find the speed from Eq,
vx = v0x = 11.4 m/s and
vy = v0y + ayt
= 0 + (–9.8 m/s2)(1.8 s)
vy = –17.4 m/s
Then the speed of the ball at impact is
v = [vx2 + vy2]1/2
v = [(11.4 m/s)2 + (−17.4 m/s)2]1/2 = 20.8 m/s
The speed of the ball when it hits the water is 20.8 m/s .
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