Question:
In the projectile motion of an object, the object reaches its maximum height where its speed is half of initial speed. Then the ratio between range and maximum height of projectile is
In the projectile motion of an object, the object reaches its maximum height where its speed is half of initial speed. Then the ratio between range and maximum height of projectile is
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At maximum height in projectile motion, the vertical velocity becomes zero and only the horizontal velocity \(u\cos\theta\) remains. Use this condition first to find the angle of projection.
Updated On: Jun 18, 2026
- \(4\sqrt{3}\)
- \(\dfrac{\sqrt{3}}{4}\)
- \(\dfrac{4}{\sqrt{3}}\)
- \(\dfrac{2}{\sqrt{3}}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use speed at maximum height.
Let the initial speed of the projectile be \[ u \] and the angle of projection be \[ \theta. \] At maximum height, the vertical component of velocity becomes zero, so the speed at maximum height is only the horizontal component: \[ u\cos\theta \] Given that this speed is half of the initial speed, \[ u\cos\theta=\frac{u}{2} \] \[ \cos\theta=\frac{1}{2} \] Therefore, \[ \theta=60^\circ \]
Step 2: Write the formula for range.
The range of a projectile is \[ R=\frac{u^2\sin 2\theta}{g} \] Substituting \[ \theta=60^\circ, \] \[ R=\frac{u^2\sin 120^\circ}{g} \] Since \[ \sin 120^\circ=\frac{\sqrt{3}}{2}, \] we get \[ R=\frac{\sqrt{3}u^2}{2g} \]
Step 3: Write the formula for maximum height.
The maximum height is \[ H=\frac{u^2\sin^2\theta}{2g} \] Substituting \[ \theta=60^\circ, \] \[ H=\frac{u^2\sin^2 60^\circ}{2g} \] Since \[ \sin 60^\circ=\frac{\sqrt{3}}{2}, \] \[ \sin^2 60^\circ=\frac{3}{4} \] Therefore, \[ H=\frac{u^2\cdot \frac{3}{4}}{2g} \] \[ H=\frac{3u^2}{8g} \]
Step 4: Find the ratio \(\dfrac{R}{H}\).
\[ \frac{R}{H} = \frac{\frac{\sqrt{3}u^2}{2g}}{\frac{3u^2}{8g}} \] \[ = \frac{\sqrt{3}u^2}{2g}\cdot \frac{8g}{3u^2} \] \[ = \frac{4\sqrt{3}}{3} \] \[ = \frac{4}{\sqrt{3}} \]
Step 5: Final conclusion.
Therefore, the ratio between range and maximum height is \[ \boxed{\frac{4}{\sqrt{3}}} \]
Let the initial speed of the projectile be \[ u \] and the angle of projection be \[ \theta. \] At maximum height, the vertical component of velocity becomes zero, so the speed at maximum height is only the horizontal component: \[ u\cos\theta \] Given that this speed is half of the initial speed, \[ u\cos\theta=\frac{u}{2} \] \[ \cos\theta=\frac{1}{2} \] Therefore, \[ \theta=60^\circ \]
Step 2: Write the formula for range.
The range of a projectile is \[ R=\frac{u^2\sin 2\theta}{g} \] Substituting \[ \theta=60^\circ, \] \[ R=\frac{u^2\sin 120^\circ}{g} \] Since \[ \sin 120^\circ=\frac{\sqrt{3}}{2}, \] we get \[ R=\frac{\sqrt{3}u^2}{2g} \]
Step 3: Write the formula for maximum height.
The maximum height is \[ H=\frac{u^2\sin^2\theta}{2g} \] Substituting \[ \theta=60^\circ, \] \[ H=\frac{u^2\sin^2 60^\circ}{2g} \] Since \[ \sin 60^\circ=\frac{\sqrt{3}}{2}, \] \[ \sin^2 60^\circ=\frac{3}{4} \] Therefore, \[ H=\frac{u^2\cdot \frac{3}{4}}{2g} \] \[ H=\frac{3u^2}{8g} \]
Step 4: Find the ratio \(\dfrac{R}{H}\).
\[ \frac{R}{H} = \frac{\frac{\sqrt{3}u^2}{2g}}{\frac{3u^2}{8g}} \] \[ = \frac{\sqrt{3}u^2}{2g}\cdot \frac{8g}{3u^2} \] \[ = \frac{4\sqrt{3}}{3} \] \[ = \frac{4}{\sqrt{3}} \]
Step 5: Final conclusion.
Therefore, the ratio between range and maximum height is \[ \boxed{\frac{4}{\sqrt{3}}} \]
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