Question

a stone is thrown vertically upward from the top of a building with the v = 30m/s . another stone is dropped 4 sec after the first is thrown up. find out when the first stone will meet the second stone'

Answer 1

Explanation:

for the first stone

v=u-gt

u=30m/s

v=0

0=30-10t

10t=30

t=3sec

second stone is dropped 4 sec after first stone

so t=7sec

u=0

s=0×7+1/2×10×7²

s=5×49=245m

the firsts syone will cover

v²=u²-2gs

0=30²-2×10×s

20s=900

S=900/20=45m

meeting point =245-45=200m from ground