Projectile Problems with Solutions
Problem#1
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 300 m/s at 55.0° above the horizontal. It explodes on the mountainside 42.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?Answer:
initial velocity component given by
vxi = vi cos θ = 300 m/s cos 55.00 = 172.07 m/s
vyi = vi sin θ = 300 m/s sin 55.00 = 245.75 m/s
then, the x and y coordinates of the shell where it explodes, relative to its firing point is
x = vxit = (172.07 m/s) x 42.0 s = 7.23 x 103 m and
y = y0 + vyit – ½ gt2 = (300 m/s)(sin 55.00)(42.0 s) – ½ (9.80 m/s2)(42.0 s)2 = 1.68 x 103 m
Problem#2
In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is
momentarily distracted and does not see the mug, which slides off the counter and strikes the floor 1.40 m from the base of the counter. If the height of the counter is 0.860 m, (a) with what velocity did the mug leave the counter, and (b) what was the direction of the mug’s velocity just before it hit the floor?
Answer:
(a) The mug leaves the counter horizontally with a velocity vxi (say). If time t elapses before it hits the ground, then since there is no horizontal acceleration, xf = vxit , i.e.,
t = xf/vxi = 1.40 m/vxi
In the same time it falls a distance of 0.860 m with acceleration downward of 9.80 m/s-2. Then
yf = yi + vyit + ½ ayt2
0 = 0.860 + 0 + ½ (–9.8 m/s2)( 1.40 m/vxi)2
vxi = 3.34 m/s
(b) The vertical velocity component with which it hits the floor is
vyf = vyi + ayt = 0 + (–9.8 m/s2)(1.4 m/3.34 m/s) = –4.11 m/s
Hence, the angle θ at which the mug strikes the floor is given by
tan θ = vy/vx
θ = tan-1(–4.11/3.34) = –50.90
Problem#3
In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor at distance d from the base of the counter. The height of the counter is h. (a) With what velocity did the mug leave the counter, and (b) what was the direction of the mug’s velocity just before it hit the floor?
Answer:
The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking
the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are
xf = vxit and yf = yi + vyit + ½ ayt2 = 0 + 0 + ½(–g)t2
When the mug reaches the floor, yf = −h so
−h = ½(–g)t2
which gives the time of impact as
t = √(2h/g)
(a) Since xf = d when the mug reaches the floor, xf = vxit becomes
vxi = d/t = d√(g/2h)
(b) Just before impact, the x-component of velocity is still
vxf = vxi’
while the y-component is
vyf = vyi + ayt
vyf = 0 – g√(2h/g)
Then the direction of motion just before impact is below the horizontal at an angle of
tan θ = vy/vx
θ = tan-1[g√(2h/g)/d√(g/2h)]
θ = tan-1(2h/g)
Problem#4
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25.0 m/s. The first one is thrown at an angle of 70.0° with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first to arrive at the same time?
Answer:
(a) The time of flight of the first snowball is the nonzero root of
yf = yi + vyit1 – ½ gt12
0 = 0 + (25.0 m/s) sin 700 t1 – ½ (9.80 m/s2)t12
t = [2 x (25 m/s) sin 700]/(9.80 m/s2) = 4.79 s
The distance to your target is
xf – xi = vxit1 = (25.0 m/s) cos 70.00 (4.79 s) = 41.0 m
Now the second snowball we describe by
yf = yi + vyit2 + ½ ay t22
0 = 0 + (25.0 m/s) sin θ2 t2 – ½ (–9.80 m/s2)t22
t2 = [2 x (25 m/s) sin θ2]/(9.80 m/s2) = 4.79 s
t2 = (5.10 s) sin θ2
The distance to your target is
xf – xi = vxit2
41.0 m – 0 = (25 m/s) cos θ2 (5.10 s) sin θ2
cos θ2 sin θ2 = 0.321
Using sin 2θ = 2 sinθ cosθ we can solve
sin 2θ2 = 0.643
2θ2 = sin-1(0.643)
θ2 = 20.00
(b) The second snowball is in the air for time
t2 = (5.10 s) sinθ2 = (5.10 s) sin20° = 1.75 s,
so you throw it after the first by
t1 – t2 = 4.79 s – 1.75 s = 3.05 s
In the same time it falls a distance of 0.860 m with acceleration downward of 9.80 m/s-2. Then
yf = yi + vyit + ½ ayt2
0 = 0.860 + 0 + ½ (–9.8 m/s2)( 1.40 m/vxi)2
vxi = 3.34 m/s
(b) The vertical velocity component with which it hits the floor is
vyf = vyi + ayt = 0 + (–9.8 m/s2)(1.4 m/3.34 m/s) = –4.11 m/s
Hence, the angle θ at which the mug strikes the floor is given by
tan θ = vy/vx
θ = tan-1(–4.11/3.34) = –50.90
Problem#3
In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor at distance d from the base of the counter. The height of the counter is h. (a) With what velocity did the mug leave the counter, and (b) what was the direction of the mug’s velocity just before it hit the floor?
Answer:
The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking
the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are
xf = vxit and yf = yi + vyit + ½ ayt2 = 0 + 0 + ½(–g)t2
When the mug reaches the floor, yf = −h so
−h = ½(–g)t2
which gives the time of impact as
t = √(2h/g)
(a) Since xf = d when the mug reaches the floor, xf = vxit becomes
vxi = d/t = d√(g/2h)
(b) Just before impact, the x-component of velocity is still
vxf = vxi’
while the y-component is
vyf = vyi + ayt
vyf = 0 – g√(2h/g)
Then the direction of motion just before impact is below the horizontal at an angle of
tan θ = vy/vx
θ = tan-1[g√(2h/g)/d√(g/2h)]
θ = tan-1(2h/g)
Problem#4
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25.0 m/s. The first one is thrown at an angle of 70.0° with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first to arrive at the same time?
Answer:
(a) The time of flight of the first snowball is the nonzero root of
yf = yi + vyit1 – ½ gt12
0 = 0 + (25.0 m/s) sin 700 t1 – ½ (9.80 m/s2)t12
t = [2 x (25 m/s) sin 700]/(9.80 m/s2) = 4.79 s
The distance to your target is
xf – xi = vxit1 = (25.0 m/s) cos 70.00 (4.79 s) = 41.0 m
Now the second snowball we describe by
yf = yi + vyit2 + ½ ay t22
0 = 0 + (25.0 m/s) sin θ2 t2 – ½ (–9.80 m/s2)t22
t2 = [2 x (25 m/s) sin θ2]/(9.80 m/s2) = 4.79 s
t2 = (5.10 s) sin θ2
The distance to your target is
xf – xi = vxit2
41.0 m – 0 = (25 m/s) cos θ2 (5.10 s) sin θ2
cos θ2 sin θ2 = 0.321
Using sin 2θ = 2 sinθ cosθ we can solve
sin 2θ2 = 0.643
2θ2 = sin-1(0.643)
θ2 = 20.00
(b) The second snowball is in the air for time
t2 = (5.10 s) sinθ2 = (5.10 s) sin20° = 1.75 s,
so you throw it after the first by
t1 – t2 = 4.79 s – 1.75 s = 3.05 s
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