Question

A particle is projected at an angle of \(60^\circ\) with the ground. When the projectile makes an angle \(45^\circ\) with the horizontal, its speed becomes \(20\,\text{m s}^{-1}\). Then the initial velocity is:

• A. \(20\sqrt{2}\,\text{m s}^{-1}\)
• B. \(10\sqrt{2}\,\text{m s}^{-1}\)
• C. \(5\sqrt{5}\,\text{m s}^{-1}\)
• D. \(10\sqrt{5}\,\text{m s}^{-1}\)

Hint:

In projectile motion:
Horizontal velocity remains constant
Angle of velocity gives ratio of vertical and horizontal components

Answer 1

The Correct Option is A

Concept:
In projectile motion:
Horizontal component of velocity remains constant
Direction of velocity at any instant depends on the ratio of vertical to horizontal components If velocity makes angle \(\theta\) with horizontal: \[ \tan\theta=\frac{v_y}{v_x} \]

Step 1: Write horizontal component of velocity. Let initial speed be \(u\). \[ v_x = u\cos 60^\circ = \frac{u}{2} \]
Step 2: Use the condition when velocity makes \(45^\circ\). At that instant: \[ \tan 45^\circ = \frac{v_y}{v_x} = 1 \Rightarrow v_y = v_x = \frac{u}{2} \]
Step 3: Use given speed at that instant. Speed: \[ v=\sqrt{v_x^2+v_y^2} = \sqrt{\left(\frac{u}{2}\right)^2+\left(\frac{u}{2}\right)^2} = \frac{u}{\sqrt{2}} \] Given \(v=20\): \[ \frac{u}{\sqrt{2}} = 20 \Rightarrow u = 20\sqrt{2} \] \[ \boxed{u = 20\sqrt{2}\,\text{m s}^{-1}} \]