Question

A body is projected at an angle of ${30}^{\circ }$ with the horizontal and with a speed of $30m{s}^{-1}$. What is the angle with the horizontal after $1.5s$? $(g=10m{s}^{-2})$.

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

Answer: (A)

$0^{\circ }$

The time of flight is given by
$T=\frac{2u\sin \theta }{g}=\frac{2\times 30\times 1/2}{10}=3s$
Thus, after $1.5s$ the body will be at the highest point.
So the direction of motion will be horizontal after $1.5s$, the angle with the horizontal is $0^{\circ }$.