A softball player is testing her physics knowledge and tosses a softball upward from the top of a building. The building is 100 m tall and the ball starts with a velocity of 10 m/s. a. What is the maximum height the ball reaches and at what time? b. How much time does it take for the ball to go from h=50 m to h=0 m on the way down? c. What is the velocity when h=50 m?

Explanation:

We shall apply newton's laws formula

a )

initial velocity in upward direction u = 10 m/s

acceleration due to gravity g = 9.8/ m .s²

Let h be the maximum height where v = o

v² = u² - 2gh

0 = 10² - 2 gh

h = 10² / 2g

= 10² /  2  x 9.8

= 5.10 m

Since the ball was thrown from height of 100 m , total maximum height of ball

= 100 + 5.10

= 105.10 m

Let t be the time taken

v = u - gt

0 = 10 - gt

t = 10 / 9.8

= 1.02 s

b )

when h = 50 on its way downwards , velocity

v² = u² + 2 g s  

v² = 0 + 2 x 9.8 x ( 105.10 - 50 )

[  distance travelled by ball at this point from top = 105.1 - 50 = 55.10 ]

v = 32.86 m / s

Let us find out final velocity of touching the ground . For it distance travelled = 105.10

v² = u² + 2gh

v² = 0 + 2 x 9.8 x 105.1

v = 45.39 m /s

Now velocity at h = 50 is 32.86

velocity at h = 0 is 45.39

time taken to travel fro h = 50 to h = 0

v = u + gt

45 .39 = 32.86 + 9.8 x t

t =  1.28 s .

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