A body is dropped from a tower of height 96 m. Another body is thrown up from the ground after a second with a velocity of 40 m/s. When and where they will meet?

SOUNDARYA asked

Answer:

For the first body,

u=0 m/s

Let S=x (from top)

x=0.5 gt^{2 } ————— (1)

For the second body,

u=40 m/s

a=-g

S=96-x

96-x=40 t -0.5 gt^{2 } ———–(2)

(1) + (2) =>

96=40t

or

t=96/40 = 2.4 s

x= 0.5 g t^{2 }=**28.224 m** from top (taking g = 9.8 m/s^{2})

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