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})