Arock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. assume free fall : part a) in how many seconds after being thrown does the rock strike the ground? part b) what is the speed of the rock just before it strikes the ground?

(a) 5.7 s

(b) 39 m/s

Explanation:

(a) u = 18 m/s

At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.

use first equation of motion.

v = u + g t

0 = 18 - 10 x t

t = 1.8 s

Let the maximum height attained by the ball when it thrown from 50 m height is h'.

Use third equation of motion

v^2 = u^2 + 2 g h'

0 = 18^2 - 2 x 10 x h'

h' = 16.2 m

Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m

Let t' be the time taken by the ball to hit the ground as it falls from maximum height.

use third equation of motion

H = ut + 1/2 x g t'^2

76.2 = 0 + 1/2 x 10 x t'^2

t' = 3.9 s

Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s

(b) Let v be the velocity with which the ball strikes the ground.

v^2 = u^2 + 2 g H

v^2 = 0 + 2 x 10 x 76.2

v = 39 m/s

this means that the temperature of both substances will be equal