Question

A body is projected horizontally with
a velocity
$\sqrt{29}$
m/s from a height
10 m, the velocity on reaching the
ground ................ m/s.
(
A)
$\sqrt{29}$
B) 10 C) 15
D) 20​

Answer 1

Answer:

15m/s

Step-by-step explanation:

$u = \sqrt{29}$

h=10m

g=9.8m/s^2

v=?

${v}^{2} - {u}^{2} = 2gh$

${v}^{2} - 29 = 2 \times 9.8 \times 10 \\ {v}^{2} = 196 + 29 \\ {v }^{2} = 225 \\ v = 15$

HOPE THIS HELPS!!!!!! :)