Question

A stone is dropped into water from a bridge 44.1 m above the water
level. Another stone is thrown vertically downward I s later. Both
stones reach the water surface simultaneously. Find the initial
downward velocity of the second stone ifg=9.8 m/s

Answer 1

x=44.1m

a=9.8m/s2

u=0

Wkt,

x=ut+1/2gt2

44.1=1/2*9.8*t2
t2=441/49

t=3 sec. So the second stone will come down in 3-1=2sec

Therefore, applying the same eqn again we get,

44.1=2u+1/2*9.8*4

44.1=2u+19.6

u=12.25m/s

it's may help you

Answer 2

Answer:

It may helps you brother