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 $1.5$ seconds ?
$(g=10m{s}^{-2}).$

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is A $0^{\circ }$

Given, $u=30m/s$ . Let $\alpha$ be the angle of projectile with the horizontal after $1.5$ seconds. Let $v$ be the velocity of the projectile after $1.5$ seconds.
$v_x=u_x=u\cos \theta \Rightarrow v_x=30\cos {30}^{\circ }=15\sqrt{3}m/s$
Using first equation of motion for the vertical component of velocity, we get
$v_y=u_y-gt\Rightarrow v_y=30\sin {30}^{\circ }-10\times 1.5=0$
Vertical component of velocity becomes $0$ when the body is at the highest point of the projectile. $\therefore \alpha =0^{\circ }$