A ball is thrown vertically upward from the top of a building with a velocity of 20m/s ,the ball reaches a max height in 2s and return back to the ground the height of the building is 25m .Calculate the distance travelled upward by the ball Calculate the velocity with which the ball lands on the ground

Distance travelled upward = 20m

Velocity when landed = 30ms^-2

Explanation:

At the maximum height , velocity = 0

To the upward motion

v^2 = u^2 + 2as

v = 0

u = 20

s= h

a = g = 10ms^-2

0 = ( 20)^2 - 2( 10 ) h

20h = 20 * 20

h = 20m

To the downward motion

v^2 = u^2 + 2as

s= 20 + 25

s = 45m

u = 0

v^2 = 0 + 2(10) ( 45 )

v^2 = 900

v = 30 ms^-1

the body's situation won't change the forces are equal opposit and on the same body

option number c is the right answer.