Answer to Question #91637 in Mechanics | Relativity for Marisol Cuevas

Question

A ball is thrown vertically upwards with initial speed 10m/s^-1 from the edge of a roof of height H= 20m. How long does it take for the ball to hit the ground? At what velocity does it hit the ground?

Answer 1

A Initially, consider the stage of throwing the ball up:

1) Determine the time of lifting the ball. At the same time at the maximum height the speed of the ball will be equal to zero. We have:

$$t = v_0/g$$

$$t = 10/10 = 1s$$

2) Determine the maximum lift height:

$$h_{max} = v_0*t - gt^2/2$$

$$h_{max} = 10*1 - 10*1^2/2 =10 - 5 = 5m$$

B Now consider the moment of falling the ball after a obtaining its maximum height:

3) The speed of the ball at the time of the fall:

$$v = \sqrt{2gh}$$

$$h = h_0 + h_{max}$$

$$v = \sqrt{2g(h_0 + h_{max}} = \sqrt{2*10*(20+5)} = \sqrt{20*25} = \sqrt{500} = 22.36m/s$$

4) Now calculate the time of the fall:

$$t' = v/g$$

$$t' = 22.36/10 = 2.24s$$

С Now consider the total time (time to climb up after a throw and the time the ball falls):

"t" = t' + t = 2.24 + 1 = 3.24s"

Answer: 3.24s; 22.36m/s.