Question

A stone is dropped from rest. What is it speed at 2 seconds?

Answer 1

Explanation:

v=u+at or v=9.8t

t= 30/10 =>3.0s is the time at which stone will reach 30 m/s velocity 

=3.0s is the time at which stone will reach 30 m/s velocity

First stone will reach s=ut+(1/2)at²

so s1

=> 1/2×10×9m

s1 =45m

(initial velocity = u = 0)

second stone was dropped after 2sec, so time for itt 1

=> 3.0−2.0s=1.0s

s 2 => 1/2×10×1² =>5.0m

so distance between two stones

D=45.0−5.0=40.400m