Question:
A player kicks a football at an angle \(30^\circ\) with the horizontal with an initial speed \(30\,\text{ms}^{-1}\). A second player standing at a distance of \(21\sqrt{3}\,\text{m}\) from the first and in the direction of kick, starts running to catch the ball, at the same instant as kicked by first player. The minimum speed of second player to catch the ball before it hits the ground is
A player kicks a football at an angle \(30^\circ\) with the horizontal with an initial speed \(30\,\text{ms}^{-1}\). A second player standing at a distance of \(21\sqrt{3}\,\text{m}\) from the first and in the direction of kick, starts running to catch the ball, at the same instant as kicked by first player. The minimum speed of second player to catch the ball before it hits the ground is
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For minimum running speed in projectile-catching problems, assume the player catches the projectile at the last possible instant, just before it hits the ground.
Updated On: Jun 26, 2026
- \(10\,\text{ms}^{-1}\)
- \(8\,\text{ms}^{-1}\)
- \(8\sqrt{3}\,\text{ms}^{-1}\)
- \(15\sqrt{3}\,\text{ms}^{-1}\)
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The Correct Option is C
Solution and Explanation
Step 1: Find the horizontal range of the football.
The range of a projectile is \[ R=\frac{u^2\sin 2\theta}{g} \] Here, \[ u=30\,\text{ms}^{-1},\quad \theta=30^\circ,\quad g=10\,\text{ms}^{-2} \] So, \[ R=\frac{30^2\sin 60^\circ}{10} \] \[ R=\frac{900\cdot \frac{\sqrt{3}}{2}}{10} \] \[ R=45\sqrt{3}\,\text{m} \]
Step 2: Find the distance the second player must run.
The second player is initially at a distance \[ 21\sqrt{3}\,\text{m} \] from the first player.
To catch the ball just before it hits the ground, he must reach the landing point.
Therefore, required running distance is \[ 45\sqrt{3}-21\sqrt{3} \] \[ =24\sqrt{3}\,\text{m} \]
Step 3: Find the time of flight of the football.
Time of flight is \[ T=\frac{2u\sin\theta}{g} \] \[ T=\frac{2(30)\sin 30^\circ}{10} \] \[ T=\frac{60\cdot \frac{1}{2}}{10} \] \[ T=3\,\text{s} \]
Step 4: Find the minimum speed of the second player.
Minimum speed is obtained when the second player catches the ball just before it hits the ground.
So, \[ v=\frac{\text{distance}}{\text{time}} \] \[ v=\frac{24\sqrt{3}}{3} \] \[ v=8\sqrt{3}\,\text{ms}^{-1} \]
Step 5: Final conclusion.
Hence, the minimum speed of the second player is \[ \boxed{8\sqrt{3}\,\text{ms}^{-1}} \]
The range of a projectile is \[ R=\frac{u^2\sin 2\theta}{g} \] Here, \[ u=30\,\text{ms}^{-1},\quad \theta=30^\circ,\quad g=10\,\text{ms}^{-2} \] So, \[ R=\frac{30^2\sin 60^\circ}{10} \] \[ R=\frac{900\cdot \frac{\sqrt{3}}{2}}{10} \] \[ R=45\sqrt{3}\,\text{m} \]
Step 2: Find the distance the second player must run.
The second player is initially at a distance \[ 21\sqrt{3}\,\text{m} \] from the first player.
To catch the ball just before it hits the ground, he must reach the landing point.
Therefore, required running distance is \[ 45\sqrt{3}-21\sqrt{3} \] \[ =24\sqrt{3}\,\text{m} \]
Step 3: Find the time of flight of the football.
Time of flight is \[ T=\frac{2u\sin\theta}{g} \] \[ T=\frac{2(30)\sin 30^\circ}{10} \] \[ T=\frac{60\cdot \frac{1}{2}}{10} \] \[ T=3\,\text{s} \]
Step 4: Find the minimum speed of the second player.
Minimum speed is obtained when the second player catches the ball just before it hits the ground.
So, \[ v=\frac{\text{distance}}{\text{time}} \] \[ v=\frac{24\sqrt{3}}{3} \] \[ v=8\sqrt{3}\,\text{ms}^{-1} \]
Step 5: Final conclusion.
Hence, the minimum speed of the second player is \[ \boxed{8\sqrt{3}\,\text{ms}^{-1}} \]
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