A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 4m/s at the floor. Next the pebble is thrown down with an initial speed of 3m/s from the same height. What is its speed at the floor?

Hint: As a first step, one could read the question well and hence note down the given points under two sets of conditions. You could apply the equation of motion to both these cases accordingly. The height covered by the pebble obtained for the first case could be substituted to get the required answer in the second case.

Formula used:
Equation of motion,
${{v}^{2}}-{{u}^{2}}=2gs$

Complete step-by-step solution:
In the question, we are given two situations. One where the pebble is released from certain heights and the final velocity is 4m/s. For the second case, initial velocity of the pebble is given and it covers the same distance as before to reach the ground. We are supposed to find the final velocity in this case.
For the first case,
Initial velocity, u =0
Final velocity, v = 4m/s
Let the distance covered during its fall be s and the acceleration on it would be that due to gravity. We have the equation of motion,
${{v}^{2}}-{{u}^{2}}=2gs$…………………………………. (1)
$\Rightarrow {{4}^{2}}-0=2\times 10\times s$
$\Rightarrow s=0.8m$
Now, for the second case,
u = 3m/s
s =0.8m
(1) would become,
${{v}^{2}}={{3}^{2}}+\left( 2\times 10\times 0.8 \right)$
$\therefore v=5m{{s}^{-1}}$
Therefore, we found the final velocity of the pebble for the second case to be 5m/s.

Note: There is certain basic knowledge you should have while dealing with problems related to one dimensional motion. For example, here, for the motion of the pebble, force due to gravity is responsible as it is under free fall. Hence, we have taken the acceleration of the pebble to be g.