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Chapter 2: Problem 68

A projectile has the same range \(R\) for two angles of projection. If \(T_{1}\) and \(T_{2}\) be the times of flight in the two cases, then \(R\) is \(\begin{array}{ll}\text { (A) } T_{1} T_{2} g & \text { (B) } \frac{T_{1} T_{2} g}{2}\end{array}\) (C) \(\left(T_{1}^{2}+T_{1}^{2}\right) g\) (D) \(\frac{T_{1}^{2}+T_{2}^{2}}{2} g\)

Short Answer

Expert verified
The range \(R\) is \(\frac{T_{1} T_{2} g}{2}\), hence the correct answer is (B).

Step by step solution

01

Consider the initial parameters of projectile motion

In a projectile motion, the range \(R\) and time of flight \(T\) can be given as: \(R = u \cos \theta (T_1 + T_2)\) where \(u\) is the initial speed and \(\theta\) is the angle of projection. And \( T_1 = 2u \sin \theta / g \) and \( T_2 = 2u \sin(90^{\circ} - \theta) / g \).
02

Write T1 + T2 in terms of u/g

Since \(T_1\) and \(T_2\) are times of flights for the same projectile, their sum can be given as: \(T_1 + T_2 = (2u/g) (\sin \theta + \sin(90^{\circ} - \theta)) = (2u/g)\). Here \(g\) is acceleration due to gravity.
03

Substitute T1 + T2 into the Range equation

Now substitute \(T_1 + T_2\) from Step 2 into the range formula from Step 1, it becomes \(R = u \cos \theta * (2u/g) = u^2*(2/g) = T_1*T_2*g/2 \).
04

Identify the correct answer

Comparing our derived expression for the range to the options given in the exercise, it is found to match with option (B).

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Most popular questions from this chapter

The maximum height of a projectile for two complementary angles of projection is \(50 \mathrm{~m}\) and \(30 \mathrm{~m}\) respectively. The initial speed of projectile is (A) \(10 \sqrt{34} \mathrm{~m} / \mathrm{s}\) (B) \(40 \mathrm{~m} / \mathrm{s}\) (C) \(20 \mathrm{~m} / \mathrm{s}\) (D) \(10 \mathrm{~m} / \mathrm{s}\)

The velocity of projection of an oblique projectile is \(\vec{v}=3 \hat{i}+2 \hat{j}\) (in \(\mathrm{ms}^{-1}\) ). The speed of the projectile at the highest point of the trajectory is (A) \(3 \mathrm{~ms}^{-1}\) (B) \(2 \mathrm{~ms}^{-1}\) (C) \(1 \mathrm{~ms}^{-1}\) (D) zero

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A man wants to reach point \(C\) on the opposite bank of river. He can swim in still water with speed \(5 \mathrm{~m} / \mathrm{s}\) and can walk on ground with \(10 \mathrm{~m} / \mathrm{s}\). Velocity of river is \(3 \mathrm{~m} / \mathrm{s}\) and given \(A B=B C=500 \mathrm{~m} .\) $$ \begin{aligned} &\text { Column-I } & \text { Column-II } \\ &\hline \text { (A) The time to reach point } C \text { if } & \text { (1) } 125 \mathrm{~s} \\ &\text { he crosses the river by shortest } \\ &\text { path is } \\ &\text { (B) The time to reach point } C \text { if } & \text { (2) } 175 \mathrm{~s} \\ &\text { he crosses the river in shortest } \\ &\text { time. } \\ &\text { (C) The time to reach point } C \text { if } & \text { (3) } 120 \mathrm{~s} \\ &\text { he crosses the river along } & \text { (4) } 106 \mathrm{~s} \\ &\text { path } A C \text { (approximate value). } & \\ &\text { (D) Time to reach point } B \\ &\text { with zero drift } \end{aligned} $$
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