Question

A ball is projected from the ground with a speed 15 ms¹ at an angle θ with horizontal so that its range and maximum height are equal, then ‘tan θ’ will be equal to

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. 2
• D. 4

Answer 1

The Correct Option is D

To solve the problem, we need to equate the range and the maximum height of the projectile. We start by understanding the formulas used for range and maximum height when a ball is projected with an initial velocity \(u\) at an angle \(\theta\) from the horizontal:

• Range (R) is given by the formula: \(R = \frac{u^2 \sin(2\theta)}{g}\)
• Maximum Height (H) is given by the formula: \(H = \frac{u^2 \sin^2(\theta)}{2g}\)

Given that the range and maximum height are equal, we have:

\(\frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{2g}\)

By cancelling common terms and simplifying, we get:

\(2 \sin(2\theta) = \sin^2(\theta)\)

Using the trigonometric identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), we substitute:

\(4 \sin(\theta)\cos(\theta) = \sin^2(\theta)\)

Divide both sides by \(\sin(\theta)\) (assuming \(\sin(\theta)\) is not zero):

\(4 \cos(\theta) = \sin(\theta)\)

This simplifies to:

\(\tan(\theta) = 4\)

Therefore, the value of \(\tan(\theta)\) is 4. Thus, the correct answer is the option

4