First slide

Projection Under uniform Acceleration

Question

A particle is projected at 60^o to the horizontal with a kinetic energy K. The kinetic energy at the highest point is

Easy

A

K/4
B

K/2
C

K
D

Zero

Solution

The kinetic energy at the highest point is given by
$\begin{array}{l} K.E. = (K.E.)_{initial} \cos^{2} α \\ = K {(\cos 600^{\circ})}^{2} = K {(\frac{1}{2})}^{2} = \frac{K}{4} \end{array}$

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