Question

A projectile is thrown upward at an angle $60^\circ$ with the horizontal. The speed of the projectile is $20$ m/s when its direction of motion is $45^\circ$ with the horizontal. The initial speed of the projectile is ________ m/s.

• A. $20\sqrt{2}$
• B. $40$
• C. $20\sqrt{3}$
• D. $40\sqrt{2}$

Hint:

In projectile motion, horizontal velocity remains constant while vertical velocity changes due to gravity. Direction $45^\circ$ implies equal velocity components.

Answer 1

The Correct Option is A

Let the initial speed of the projectile be $u$.
Step 1: Resolve initial velocity components.
Horizontal component: \[ u_x = u\cos60^\circ = \frac{u}{2} \] Vertical component: \[ u_y = u\sin60^\circ = \frac{\sqrt{3}u}{2} \] Step 2: Use the condition when direction is $45^\circ$.
At the instant when the direction of velocity is $45^\circ$, \[ \frac{v_y}{v_x} = \tan45^\circ = 1 \Rightarrow v_y = v_x \] Since horizontal velocity remains constant: \[ v_x = \frac{u}{2} \Rightarrow v_y = \frac{u}{2} \] Step 3: Use the given speed at that instant.
Speed is given as $20$ m/s: \[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{\left(\frac{u}{2}\right)^2 + \left(\frac{u}{2}\right)^2} = \frac{u}{\sqrt{2}} \] \[ \frac{u}{\sqrt{2}} = 20 \Rightarrow u = 20\sqrt{2} \] Final Answer: $\boxed{20\sqrt{2}}$