Question

a stone is dropped into a well of 20 metre deep another stone is thrown downward with velocity V ,one second later if both stones reach the water surface in the Well simultaneously V is equal to
g=10m/s^2
a.30m/s
b.15m/s
c.20m/s
d.10m/s​

Answer 1

The time taken by the first stone to fall (in air to the top of the well) 20 m below the drop point is given by the relation,

s = u t + ½ g t².

In our case u= 0 m/s (as the stone is just dropped), s= 20 m, taking g= 10 m/s², we get from the relation,

20 m = 0×t +½ ×10 m/s² × t² s². Solving we get t= 2 s.

It would takes the first stone 2s to fall 20m to the top of the water in the well.

The second stone needs to cover these 20m in 1 s ( as it is thrown with a velocity v after the first stone is dropped). Using the relation, and substituting various values we get,

20 m = v m/s × 1s +½ ×10 m/s² ×1 s²= v + 5 m

So,

v = 15 m/s.

The second stone needs to be thrown with a velocity 15 m/

Answer 2

Answer:

hey mate your answer is 15ms^-1