Question:
A cricketer can throw a ball with a speed of \(V_0\). If he throws the ball while running with a speed \(V_0\), at an angle \(\alpha\) with the horizontal, then the range of the ball will be maximum when \(\alpha\) is
A cricketer can throw a ball with a speed of \(V_0\). If he throws the ball while running with a speed \(V_0\), at an angle \(\alpha\) with the horizontal, then the range of the ball will be maximum when \(\alpha\) is
Show Hint
When a projectile is launched from a moving platform, first add the platform velocity to the projectile velocity vector. Then use
\[
R=u_x\left(\frac{2u_y}{g}\right)
\]
and maximize the resulting expression.
Updated On: Jul 29, 2026
- \(45^\circ\)
- \(30^\circ\)
- \(60^\circ\)
- \(53^\circ\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
The velocity of projection relative to the ground is the vector sum of:
\[
\text{running velocity}
=
V_0
\]
and
\[
\text{throwing velocity}
=
V_0.
\]
The range is maximized by maximizing the projectile range expression with the resultant horizontal and vertical components.
Step 1: Find the horizontal and vertical components of the projectile velocity. Horizontal component: \[ u_x = V_0+V_0\cos\alpha = V_0(1+\cos\alpha). \] Vertical component: \[ u_y = V_0\sin\alpha. \]
Step 2: Write the expression for range. For a projectile, \[ R = u_x \left(\frac{2u_y}{g}\right). \] Substituting, \[ R = \frac{2V_0^2}{g} (1+\cos\alpha)\sin\alpha. \] Using \[ 1+\cos\alpha = 2\cos^2\frac{\alpha}{2}, \] and \[ \sin\alpha = 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}, \] \[ R = \frac{8V_0^2}{g} \sin\frac{\alpha}{2} \cos^3\frac{\alpha}{2}. \]
Step 3: Maximize the function. Let \[ x=\frac{\alpha}{2}. \] Then maximize \[ f(x)=\sin x\,\cos^3x. \] Differentiating, \[ f'(x) = \cos^4x - 3\sin^2x\cos^2x. \] \[ = \cos^2x \left( \cos^2x-3\sin^2x \right). \] For maximum, \[ \cos^2x=3\sin^2x. \] \[ \tan^2x=\frac13. \] \[ \tan x=\frac1{\sqrt3}. \] \[ x=30^\circ. \] Hence, \[ \alpha=2x=60^\circ. \] Therefore, \[ \boxed{\alpha=60^\circ} \] \[ \boxed{\text{Answer = (C)}} \]
Step 1: Find the horizontal and vertical components of the projectile velocity. Horizontal component: \[ u_x = V_0+V_0\cos\alpha = V_0(1+\cos\alpha). \] Vertical component: \[ u_y = V_0\sin\alpha. \]
Step 2: Write the expression for range. For a projectile, \[ R = u_x \left(\frac{2u_y}{g}\right). \] Substituting, \[ R = \frac{2V_0^2}{g} (1+\cos\alpha)\sin\alpha. \] Using \[ 1+\cos\alpha = 2\cos^2\frac{\alpha}{2}, \] and \[ \sin\alpha = 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}, \] \[ R = \frac{8V_0^2}{g} \sin\frac{\alpha}{2} \cos^3\frac{\alpha}{2}. \]
Step 3: Maximize the function. Let \[ x=\frac{\alpha}{2}. \] Then maximize \[ f(x)=\sin x\,\cos^3x. \] Differentiating, \[ f'(x) = \cos^4x - 3\sin^2x\cos^2x. \] \[ = \cos^2x \left( \cos^2x-3\sin^2x \right). \] For maximum, \[ \cos^2x=3\sin^2x. \] \[ \tan^2x=\frac13. \] \[ \tan x=\frac1{\sqrt3}. \] \[ x=30^\circ. \] Hence, \[ \alpha=2x=60^\circ. \] Therefore, \[ \boxed{\alpha=60^\circ} \] \[ \boxed{\text{Answer = (C)}} \]
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