A stone is thrown straight upward with a speed of 15ms. It is caught on its way down at a point 2point 5m above where it was thrown. i How fast was it going when it was caught how long does it take?

Answer 1

easy

for upward path

applying eqn

v^2 - u^2 = 2aS

and v =0, a = -g = -9.8

we get s = 11.479

now on way down it was cought on 2.5 m above ground so

s for downward = 9.479

so as u for downward = 0, a = g = 9.8 , s = 9.479

applying v^2 - u^2 = 2aS again we get

v = 13.63 mps

now as v = u +at

therf. t = 1.39 :)

by ravinder balhara. :)