Question

A projectile is thrown in the upward direction making an angle of 60° with the horizontal direction, with a velocity of 147ms$^{-1}$ Then the time after which inclination with the horizontal is 45°, is

• A. 15($\sqrt{3}$ - 1)s
• B. 15($\sqrt{3}$ + 1)s
• C. 7.5($\sqrt{3}$ - 1)s
• D. 7.5($\sqrt{3}$ + 1)s

Answer 1

Answer: (C)

7.5($\sqrt{3}$ - 1)s

Answer 2

Let the initial velocity of the projectile be $u = 147 \text{ ms}^{-1}$, and the initial angle of projection with the horizontal be $\theta_0 = 60^\circ$.

The horizontal component of the initial velocity is $u_x = u \cos \theta_0 = 147 \cos 60^\circ = 147 \times \frac{1}{2} = 73.5 \text{ ms}^{-1}$.

The vertical component of the initial velocity is $u_y = u \sin \theta_0 = 147 \sin 60^\circ = 147 \times \frac{\sqrt{3}}{2} = 73.5 \sqrt{3} \text{ ms}^{-1}$.

In projectile motion, the horizontal component of velocity remains constant (neglecting air resistance), so the horizontal velocity at time $t$ is $v_x(t) = u_x = 73.5 \text{ ms}^{-1}$.

The vertical component of velocity changes due to gravity. Assuming the acceleration due to gravity is $g$, the vertical velocity at time $t$ is $v_y(t) = u_y - gt = 73.5 \sqrt{3} - gt$.

The inclination of the velocity vector with the horizontal at time $t$ is given by the angle $\theta$ such that $\tan \theta = \frac{v_y(t)}{v_x(t)}$.

We are looking for the time $t$ when the inclination with the horizontal is $\theta_f = 45^\circ$. So, $\tan 45^\circ = \frac{v_y(t)}{v_x(t)}$. Since $\tan 45^\circ = 1$, we have $v_y(t) = v_x(t)$.

Substitute the expressions for $v_x(t)$ and $v_y(t)$: $73.5 \sqrt{3} - gt = 73.5$. Let's assume $g = 9.8 \text{ ms}^{-2}$.

Then the equation is $73.5 \sqrt{3} - 9.8t = 73.5$. $9.8t = 73.5 \sqrt{3} - 73.5$ $9.8t = 73.5 (\sqrt{3} - 1)$. $t = \frac{73.5}{9.8} (\sqrt{3} - 1)$. We calculate the ratio $\frac{73.5}{9.8}$: $\frac{73.5}{9.8} = \frac{735/10}{98/10} = \frac{735}{98} = \frac{15}{2} = 7.5$.

Therefore, $t = 7.5 (\sqrt{3} - 1)$ seconds.