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°.