Projectile Problems with Solutions 4

 Problem#1

A soccer player kicks a rock horizontally off a 40.0 m high  cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

Answer:
The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from

y_f = y_i + v_yit + ½ a_yt²

0 = 40.0 m + 0 + ½ (–9.80 m/s²)t²

t = 2.86 s

The extra time 3.00 s – 2.86 s = 0.143 s is the time required for the sound she hears to travel straight back to the player. It covers distance

(343 m/s)(0.143 s) = √[x² + (40.0 m)²]

49.0 m = √[x² + (40.0 m)²]

x = 28.3 m

where x represents the horizontal distance the rock travels.

x_f = x_i + v_xit
28.3 m = 0 + v_xi(2.86 s)
Then
v_xi = 9.91 m/s

Problem#2
A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. 1). His motion through space can be modeled precisely as that of a particle at his center of mass, which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor, and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his “hang time”), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations y_i = 1.20 m, y_max = 2.50 m, y_f = 0.700 m.

Fig.1

Answer:
From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His vertical velocity and vertical displacement are related by the equation

v_yf² = v_yi² + 2a_y(y_f - y_i)

0² = v_yi² + 2(–9.80 m/s²)(1.85 m – 1.02 m)

From this, v_yi = 4.03 m/s

For the downward part of the flight, the equation gives

v_yf² = 0 + 2(–9.80 m/s²)(0.900 m – 1.85 m)

Thus the vertical velocity just before he lands is

v_yf = –4.32 m/s

(a) His hang time may then be found from

v_yf = v_yi + (–g)t

–4.32 m/s = 4.03 m/s + (–9.80 m/s²)t

t = 0.852 s

(b) Looking at the total horizontal displacement during the leap,

x_f = x_i + v_xt, becomes

2.80 m = v_xi(0.852 s)

v_xi = 3.29 m/s

(c) vertical velocity components at the instant of takeoff, v_yi = 4.03 m/s. See above for proof.

(d) The takeoff angle is:

tan θ = (v_yi/v_xi)

tan θ = (4.03/3.29)

θ = tan^-1(4.03/3.29) = 50.8⁰

(e) Similarly for the deer, the upward part of the flight gives

v_yf² = v_yi² + 2a_y (y_f - y_i)

0 = v_yi² + 2(–9.80 m/s²)(2.50 m – 1.20 m)

v_yi = 5.04 m/s

For the downward part
v_yf² = v_yi² + 2a_y (y_f - y_i)

v_yf² = 0 + 2(–9.80 m/s²)(0.700 m – 2.50 m)

v_yf = –5.94 m/s

The hang time is then found as
v_yf = v_yi + a_yt

–5.94 m/s = 5.04 m/s + (–9.80 m/s²)t

t = 1.12 s

Problem#3
An archer shoots an arrow with a velocity of 45.0 m/s at an  angle of 50.0° with the horizontal. An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. (a) What is the initial speed of the apple? (b) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?

Answer:
The arrow’s flight time to the collision point is

x_f = x_i + v_xt

150 m = 0 + (45 m/s)t

t = 5.19 s
The arrow’s altitude at the collision is

y_f = y_i + v_yit + ½ a_yt²

y_f = 0 + (45 m/s)(sin 50⁰)(5.19 s) + ½ (–9.80 m/s²)(5.19 s)

y_f = 47.0 m

(a) The required launch speed for the apple is given by

v_yf² = v_yi² + 2a_y (y_f - y_i)

0 = v_yi² + 2(–9.80 m/s²)(47.0 m – 0)

v_yi = 30.3 m/s

(b) The time of flight of the apple is given by

v_yf = v_yi + a_yt

0 = 30.3 m/s + (–9.80 m/s²)t

t = 3.10 s

So the apple should be launched after the arrow by

5.19 s – 3.10 s = 2.09 s

Problem#4
A fireworks rocket explodes at height h, the peak of its vertical trajectory. It throws out burning fragments in all directions, but all at the same speed v. Pellets of solidified metal fall to the ground without air resistance. Find the smallest angle that the final velocity of an impacting fragment makes with the horizontal.

Answer:
For the smallest impact angle

tan θ_i = (v_yf/v_yi)

we want to minimize v_yf and maximize v_xf = v_xi.

The final y-component of velocity is related to v_yi by

v_yf² = v_yi² + 2a_y (y_f - y_i) = v_yi² + 2gh

so we want to minimize v_yi and maximize v_xi . Both are accomplished by making the initial velocity
horizontal. Then v_xi = v, v_yi = 0, and

v_yf = √(2gh)

At last, the impact angle is

θ_i = tan^-1 [√(2gh)/v]