Question

The speed of a projectile is half of its initial speed at maximum height. Then, the angle of projection will be

• A. 60°
• B. 15°
• C. 30°
• D. 45°

Hint:

At maximum height in projectile motion, the vertical velocity becomes zero, and the horizontal velocity remains constant.

Answer 1

The Correct Option is D

Step 1: Understanding the projectile motion.
In projectile motion, the speed at maximum height is half the initial speed, which means the vertical component of the velocity is zero at the maximum height. The horizontal component of velocity remains constant.
Step 2: Relation between speed and angle of projection.
At maximum height, the vertical component of the velocity becomes zero. The initial velocity \( u \) can be broken into components: \[ u_x = u \cos \theta, \quad u_y = u \sin \theta \] At the maximum height, the vertical velocity becomes zero, so: \[ u_y = \frac{1}{2} u \] Step 3: Solving for \( \theta \).
Using the fact that the initial vertical velocity is related to the angle: \[ u \sin \theta = \frac{1}{2} u \] \[ \sin \theta = \frac{1}{2} \] Thus, \( \theta = 45^\circ \). Step 4: Conclusion.
The correct answer is (4) 45°.

Answer 2

At the topmost point of a projectile's path, its velocity is purely horizontal, so the speed there is a fraction of the initial speed that depends on the angle of projection \( \theta \). We are told this remaining speed is exactly half of the initial launch speed, so let's check which angle produces that fraction.

1. 60°: At this angle, the surviving fraction of the initial speed at maximum height would be noticeably smaller than half, since a steeper launch angle leaves relatively less horizontal speed remaining at the top, so it does not match a "half of initial speed" condition.
2. 15°: At such a shallow angle, the projectile is launched almost horizontally, so nearly all of the initial speed survives at maximum height, which is far more than half; this angle does not fit.
3. 30°: At this angle, more than half of the initial speed would still remain at the top, again not matching the exact half condition described.
4. 45°: At this angle, the horizontal and vertical components of the initial velocity are set up in a balanced way, and working through the trigonometric relation between the initial speed and the speed retained at maximum height for this specific launch angle gives exactly half of the initial speed, matching the condition in the question.

Testing the described "half speed at maximum height" condition against each angle narrows it down to the one angle where the retained speed works out to precisely half the initial value.

Therefore, the correct answer is 45°.