Question

A man drops a ball downside from the roof of tower of height 400 m at the same time another ball is thrown upside with velocity 50 m/sec from the surface of tower then they will meet at which height from the surface of tower

Answer 1

For meeting of two bodies time travelled by the two bodies are equal.
Let that time be' t'.
#For freely falling body S1= 1/2(g)t^2.
#For upward moving body distance travelled S2 =ut +1/2(-g)t^2.
by putting S1 in S2
S2=ut - S1
S1+S2=ut .》》》》》》》
#For meeting at a point both combinely
covers total height
then 400=S1+S2》》》》》》
By solving two equations
ut =400. u is given as 50m/sec
t=400/50
=8 sec .
substituting t value in second equation
S2= 50×8 -(1/2×10×8×8)
=400-320
=60 metres.
Thus they meet at 60 metres from ground.