Question

a man is standing at the top of a 60m high tower. He throws a ball vertically upwards with velocity 20m/s. After what time will the ball pass him going downward?How long after it release will the ball reach the ground?(take g=10m/s square)

Answer 1

hello friend....!!

•The ball is thrown upward,

from equations of motion we can consider,

v = u - at ( thrown up against gravitation)

here in this case V = 0 , U = 20m/s ,
a = g = 10 m/s^2

therefore ,

0 = 20 - 10t

t = 2 seconds. ( when thrown up)

so to cross him the timetaken by the ball might be,

t = 2 + 2 = 4 seconds.

• in case (2) we should calculate the time taken by the ball to reach the ground,

we know,

$s \: \: = ut \: - \frac{1}{2} a {t}^{2}$

s = 20 X 2 - 1/2 (10)(4)

S = 40 - 20

s = 20 m

given height of the tower is 60 m

so the total height is

S = 60 + 20 = 80 m

now to calculate the time taken,

S = 1/2at^2

80 = 1/2 (10)t^2

therefore,

$t \: \: = \sqrt{16}$

therefore,

T = 4 seconds.


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hope it helps.. !!!