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Pseudo Force

A body is pro...

Question

A body is projected such that its K.E. at the top is $(3 / 4)^{t h}$ of its initial K.E. What is the angle of projectile with the vertical?

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Solution

$K E_{t o p} = \frac{3}{4} K E_{i n i t i a l}$
$\frac{1}{2} m (v c o s θ)^{2} = \frac{3}{4} \frac{1}{2} m v^{2}$
$c o s θ = \frac{\sqrt{3}}{2}$
$θ = 30^{0}$
Angle made with the horizontal is $30^{0}$ and with vertical is $60^{0}$

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