Question

A body is projected horizontally with a velocity of $u$ $m/s$ at an angle $\beta$ with the horizontal. The kinetic energy at the highest point is $( \frac{ 3}{ 4} )^{th}$ of the initial kinetic energy. The value of $\beta$ is

• A. $30^\circ$
• B. $45^\circ$
• C. $60^\circ$
• D. $120^\circ$

Answer 1

Answer: (A)

$30^\circ$

Answer 2

The kinetic energy at the highest point would be equal to $\frac{1}{2} m(u \cos \beta)^{2}$ as the vertical component of the velocity is zero.

The initial kinetic energy is the maximum kinetic energy.
So, $KE =K \cos ^{2} \beta$
Thus, $K \cos ^{2} \beta=\frac{3}{4} K$
$\Rightarrow \cos \beta=\frac{\sqrt{3}}{2}$
$\Rightarrow \beta=30^{\circ}$