Question

When a body is thrown up at an
angle of 45° with a velocity of
100 m/s, it describes a parabola. Its
velocity on point of return down
will be
(a) zero
(b) 50 m/s
(c) *cm/s
(d) 100v2 m/s​

Answer 1

Answer:

answer is

$50 \sqrt{2} \: m \: per \: second$

Explanation:

the velocity on point of return will be the velocity at the maximum height

but at the highest point only the constant horizontal velocity

$u \cos( \theta)$

acts and no vertical velocity acts due to gravity

therefore

$u \cos( \theta) = 100( \sin(45) ) = 50 \sqrt{2}$