Question

A body is projected such that its KE at the top is 3/4th of its
initial KE. What is the angle of projectile with the horizontal?
(a) 30º (b) 60º (c) 45º (d) 120º

Answer 1

Answer:

correct option is a

Explanation:

let the initial KE be,

$(k)initial = \frac{1}{2} m {v}^{2}$

KE at the top will be provided by only the horizontal component of velocity(v cosx), because, there is no vertical component at the highest point.hence,KE at the top will be,

$(k)top = \frac{1}{2} m {(vcosx)}^{2}$

it is given that

$(k)top = \frac{3}{2} (k)initial \\ \frac{1}{2} m {v}^{2} {cos}^{2} x = \frac{3}{4} \times \frac{1}{2} m {v}^{2} \\ {cos}^{2} x = \frac{3}{4} \\ cosx = \sqrt{ \frac{3}{4} } \\ cosx = \frac{ \sqrt{3} }{2} \\ x = 30$

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