Question

A Ball is thrown vertically upward
with velocity "v"& such that it
rises for T seconds (T>1)What
is distance traversed by ball
during last one second ?​

Answer 1

Answer. Initial velocity u= v

Final velocity v=0

u=v,v=0

g=9.8m/s^2(downwards)

g=-9.8m/s^2(upward)

2gs=v^2-u^2

2*(-9.8)*s=(0)^2-(0)^2

-19.6s=0

s=19.6/0=19.6

s=19.6m

When ball return to the surface of the earth s=0 v=u+at

0=0+9.8t0=9.8t

t=9.8/0=9.8

t=9.8s

Total time taken is 9.8+9.8=19.6s