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Chapter 2: Problem 98

The maximum height attained by a projectile is increased by \(5 \%\), keeping the angle of projection constant. The corresponding percentage increase in horizontal range will be (A) \(5 \%\) (B) \(10 \%\) (C) \(15 \%\) (D) \(20 \%\)

Short Answer

Expert verified
The corresponding percentage increase in horizontal range when the maximum height attained by a projectile is increased by $5 \%$ while keeping the angle of projection constant is (A) \(5 \% \).

Step by step solution

01

Understanding the Projectile Motion

For a projectile, the maximum height and horizontal range are given by the following formulas: \(H = \frac{u^2\sin^2\theta}{2g}\) and \(R = \frac{u^2\sin2\theta}{g}\) where: - \(H\) is the maximum height - \(u\) is the initial velocity - \(\theta\) is the angle of projection - \(R\) is the horizontal range - \(g\) is the acceleration due to gravity
02

Analyzing the Effect of Increase in Maximum Height

If the projectile's maximum height is increased by 5%, the new equation for maximum height becomes: \(H' = \frac{u'^2\sin^2\theta}{2g} = 1.05H\) Since the angle of projection remains constant, we have: \(\frac{u'^2\sin^2\theta}{2g} = 1.05\frac{u^2\sin^2\theta}{2g}\)
03

Finding the Ratio of New Initial Velocity to Old Initial Velocity

From the previous step, we can write: \(u'^2 = 1.05u^2\) Taking the square root of both sides, we get: \(u' = u\sqrt{1.05}\) This gives us the ratio of the new initial velocity to the old initial velocity.
04

Calculating the New Horizontal Range

Now, let's find the new horizontal range using the new initial velocity: \(R' = \frac{u'^2\sin2\theta}{g} = \frac{(u\sqrt{1.05})^2\sin2\theta}{g} = \frac{1.05u^2\sin2\theta}{g}\)
05

Finding the Percentage Increase in Horizontal Range

Now, we need to compare the new horizontal range with the old horizontal range to find the percentage increase: \(\frac{R'}{R} = \frac{1.05u^2\sin2\theta}{g} \times \frac{g}{u^2\sin2\theta} = 1.05\) This means that the horizontal range has increased by 5% as well. So, the correct answer is: (A) \(5 \%\)

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Most popular questions from this chapter

From a canon mounted on a wagon at height \(H\) from ground, a shell is fired horizontally with a velocity \(v_{0}\) with respect to canon. The canon and wagon has combined mass \(M\) and can move freely on the horizontal surface. The horizontal distance between shell and canon when the shell touches the ground is (A) \(v_{0} \sqrt{\frac{2 H}{g}}\) (B) \(\frac{v_{0} m}{M+m} \sqrt{\frac{2 H}{g}}\) (C) \(\frac{v_{0} M}{M+m} \sqrt{\frac{2 H}{g}}\) (D) \(\frac{v_{0} m}{M} \sqrt{\frac{2 H}{g}}\)

A particle is projected at \(60^{\circ}\) to the horizontal with kinetic energy \(K\). The kinetic energy at the highest point is [2007] (A) \(K / 2\) (B) \(K\) (C) Zero (D) \(K / 4\)

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