Question

A ball 'A' is thrown upwards from ground level with a velocity of 20 ms. Another ball 'B' is thrown
downwards from a height of 40 m with the same speed. Find the height at which the two balls meet.
(take g = 10 ms?)

Pls give the answer in a written paper​

Answer 1

Answer:

total height= height of ball A + height of ball B

40=S(A)+S(B)

40=U(A)t+1/2(-g)t²+U(B)t+1/2(g)t²

40=20t-5t²+20t+5t²

40=40t

t=1sec

At 1 sec two balls meet,so

s=ut+1/2gt²

s=20(1)+1/2(10)(1)²

s=20+5

s=25m