Question:
The ranges and heights for two projectiles projected with the same initial velocity at angles \(42^\circ\) and \(48^\circ\) with the horizontal are \(R_1, R_2\) and \(H_1, H_2\) respectively. Choose the correct option :
The ranges and heights for two projectiles projected with the same initial velocity at angles \(42^\circ\) and \(48^\circ\) with the horizontal are \(R_1, R_2\) and \(H_1, H_2\) respectively. Choose the correct option :
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Angles that add to \(90^\circ\) give equal ranges; the maximum height depends on \(\sin^2\theta\), which is larger for the larger angle.
Updated On: Nov 4, 2026
- \(R_1 \gt R_2\) and \(H_1 = H_2\)
- \(R_1 = R_2\) and \(H_1 \lt H_2\)
- \(R_1 \lt R_2\) and \(H_1 \lt H_2\)
- \(R_1 = R_2\) and \(H_1 = H_2\)
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The Correct Option is B
Solution and Explanation
Range \(R = \frac{u^2\sin 2\theta}{g}\). For \(\theta_1 = 42^\circ\): \(\sin 84^\circ\). For \(\theta_2 = 48^\circ\): \(\sin 96^\circ = \sin(180^\circ - 84^\circ) = \sin 84^\circ\). Hence \(R_1 = R_2\).
Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\). Since \(\sin 42^\circ \lt \sin 48^\circ\), \(H_1 \lt H_2\).
So \(R_1 = R_2\) and \(H_1 \lt H_2\).
Answer: (2) \(R_1 = R_2\) and \(H_1 \lt H_2\)
Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\). Since \(\sin 42^\circ \lt \sin 48^\circ\), \(H_1 \lt H_2\).
So \(R_1 = R_2\) and \(H_1 \lt H_2\).
Answer: (2) \(R_1 = R_2\) and \(H_1 \lt H_2\)
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