Question

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

Answer 1

v=4

u=3

a=9.8

s=?

formula

(v)2-(u)2=2a.

(4)2-(3)2=2(9.8)s

16-9=95.74s

5=95.74s

s=5/95.74

s=500/9574

s=0.0522....

Answer 2

Answer: 5m/s

Explanation: for first pebble

u=0

v=4m/s

From equation of motion

V2-u2=2gs

16-0=2×10×s

s=0.8m

For second pebble

Height is Same I. E. s is same.... So

u=3m/s

V2 =u2+2gs

=9+2×10×0.8

=25

So, v=5m/s

I hope u will understand it nd it will help u