Question

An object is projected in the air with initial velocity u at an angle θ. The projectile motion is such that the horizontal range R, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be ______ degree.

Answer 1

Correct Answer: 15

To solve this problem, we begin by recalling the formula for the horizontal range \( R \) of a projectile: \( R = \frac{u^2 \sin(2\theta)}{g} \). Here \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Step 1: Maximum Range Condition 
The range \( R \) is maximal when the angle \( \theta \) is 45 degrees, since \( \sin(2\theta) \) becomes 1 at \( \theta = 45^\circ \). Thus, \( R_{\text{max}} = \frac{u^2}{g} \).

Step 2: Second Object's Range
We are given that the second object's range \( R' \) is half of the maximum range: \( R' = \frac{R_{\text{max}}}{2} = \frac{u^2}{2g} \).

Step 3: Find the Angle for Half Range
Now, use the range formula for the second object: \( R' = \frac{u^2 \sin(2\theta')}{g} \). Setting this equal to \( \frac{u^2}{2g} \), we have: \( \frac{u^2 \sin(2\theta')}{g} = \frac{u^2}{2g} \). Simplifying gives \( \sin(2\theta') = \frac{1}{2} \).

Step 4: Calculate \( \theta' \)
The equation \( \sin(2\theta') = \frac{1}{2} \) gives \( 2\theta' = 30^\circ \) or \( 2\theta' = 150^\circ \), resulting in \( \theta' = 15^\circ \) or \( \theta' = 75^\circ \). Since we want the smaller angle of projection for a smaller range, choose \( \theta' = 15^\circ \).

Therefore, the angle of projection for the second object is 15 degrees, which falls within the given range of 15,15.

Answer 2

\(\theta = 45^\circ\)
\(R_1=\frac{R}{2}\)
\(\frac{u^2 \sin^2 \theta_1}{g} = \frac{u^2 \sin(90^\circ)}{2g}\)
\(2\theta_1 = 30^\circ\)
\(\theta_1 = 15^\circ\)
So, the answer is 15°