Question:
A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t\(_1\) and t\(_2\) are the values of the time taken by it to hit the target in two possible ways, the product t\(_1\)t\(_2\) is -
A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t\(_1\) and t\(_2\) are the values of the time taken by it to hit the target in two possible ways, the product t\(_1\)t\(_2\) is -
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The two possible angles are \(\theta\) and \(90^\circ-\theta\); write both times of flight and multiply, using \(R = u^2\sin2\theta/g\).
Updated On: Nov 5, 2026
- \(\dfrac{2R}{g}\)
- \(\dfrac{R}{g}\)
- \(\dfrac{R}{2g}\)
- \(\dfrac{R}{4g}\)
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The Correct Option is A
Solution and Explanation
For the same speed u and the same range R, the two launch angles are \(\theta\) and \(90^\circ - \theta\).
\(t_1 = \dfrac{2u\sin\theta}{g}\), \(t_2 = \dfrac{2u\sin(90^\circ-\theta)}{g} = \dfrac{2u\cos\theta}{g}\).
\(t_1t_2 = \dfrac{4u^2\sin\theta\cos\theta}{g^2} = \dfrac{2u^2\sin2\theta}{g^2}\).
Since \(R = \dfrac{u^2\sin2\theta}{g}\), \(t_1t_2 = \dfrac{2R}{g}\).
Answer: (1) \(\dfrac{2R}{g}\)
\(t_1 = \dfrac{2u\sin\theta}{g}\), \(t_2 = \dfrac{2u\sin(90^\circ-\theta)}{g} = \dfrac{2u\cos\theta}{g}\).
\(t_1t_2 = \dfrac{4u^2\sin\theta\cos\theta}{g^2} = \dfrac{2u^2\sin2\theta}{g^2}\).
Since \(R = \dfrac{u^2\sin2\theta}{g}\), \(t_1t_2 = \dfrac{2R}{g}\).
Answer: (1) \(\dfrac{2R}{g}\)
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