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A ball is projected from the bottom of a...

A ball is projected from the bottom of an inclined plane of inclination `30^@`, with a velocity of `30 ms^(-1)`, at an angle of `30^@` with the inclined plane. If `g = 10 ms^(-2)` , then the range of the ball on given inclined plane is

A

12 m

B

60 m

C

120 m

D

600 m

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To solve the problem of finding the range of a ball projected from the bottom of an inclined plane, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Values**: - Angle of inclination of the plane, \( \theta = 30^\circ \) - Initial velocity of the ball, \( v = 30 \, \text{m/s} \) - Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) - Angle of projection with respect to the inclined plane, \( \alpha = 30^\circ \) 2. **Resolve the Initial Velocity**: - The component of the initial velocity along the inclined plane is given by: \[ v_{\text{along}} = v \cos \alpha = 30 \cos 30^\circ = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s} \] 3. **Determine the Vertical Component of Gravity**: - The effective acceleration due to gravity acting down the incline is: \[ g_{\text{effective}} = g \sin \theta = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s}^2 \] 4. **Use the Range Formula for Projectile Motion on an Incline**: - The range \( R \) on an inclined plane can be calculated using the formula: \[ R = \frac{v_{\text{along}}^2}{g_{\text{effective}}} \sin(2\alpha) \] - Here, \( \sin(2\alpha) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \). 5. **Substituting the Values**: - Substitute \( v_{\text{along}} \) and \( g_{\text{effective}} \) into the range formula: \[ R = \frac{(15\sqrt{3})^2}{5} \cdot \frac{\sqrt{3}}{2} \] - Calculate \( (15\sqrt{3})^2 = 675 \): \[ R = \frac{675}{5} \cdot \frac{\sqrt{3}}{2} = 135 \cdot \frac{\sqrt{3}}{2} = 67.5\sqrt{3} \, \text{m} \] 6. **Final Calculation**: - Approximating \( \sqrt{3} \approx 1.732 \): \[ R \approx 67.5 \times 1.732 \approx 116.5 \, \text{m} \] ### Conclusion: The range of the ball on the inclined plane is approximately \( 116.5 \, \text{m} \).
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