Question

A motor cycle stuntman drives off a cliff at 80 mi/hr and hits the safety net 2.9 s later. How high was the cliff and how far is the net from the base of the cliff?

A rock is thrown horizontally from the 2nd floor building and it lands 7.5 m from the base of the building. Assume the initial height of the rock is 5.5 m. What is the initial speed of the rock?

Answer 1

(2) Rock thrown horizontally

GIVEN:

Height = d_y = 5.5 m

Range = d_x = 7.5 m

Since the ball is thrown horizontally, then the initial speed is the same as the horizontal speed, v_x .

We know that v_x = d_x / t_air. Where t_air is the amount of time that the ball is in the air. But how do we calculate t_air?

Objects thrown horizontally will spend the same amount of time in the air as objects that are in free-fall from the same height, d_y.

So, we can calculate the time in the air using d_y = v_yt_air + ½gt²_air , where v_y = 0 m/s and thus d_y = ½gt²_air . Solving for t:

t²_air = (2d_y / g), plug in your given values ⇒ t²_air = (2)(5.5m)/(9.8m/s²) = 1.12 s²

t_air = 1.06 s

Substitute into v_x = d_x/t_air = (7.5m)/(1.06s) = 7.08 m/s

So initial speed is 7.08 m/s

(1) Initial velocity, v_x = 80 mi/hr = 35.8 m/s ; t_air = 2.9 s

Height = d_y

Range = d_x

d_y = v_yt_air + ½gt²_air where v_y = 0 m/s and thus d_y = ½gt²_air = ½(9.8)(2.9)² = 41.2 m

So height of cliff is 41.2 m

We know that v_x = d_x / t_air

from the above d_x = v_x t_air = 35.8*2.9 = 103.8 m

So th net is 103.8 m away from the cliff (which is range)