Question

Maximum height up to which a projectile can reach, if it is thrown upwards with initial speed \( u \) at an angle \( \theta \) with the ground surface, is

• A. \( \dfrac{u^2 \sin^2\theta}{2g} \)
• B. \( \dfrac{u^2 \cos^2\theta}{2g} \)
• C. \( \dfrac{u \sin\theta}{2g} \)
• D. \( \dfrac{u \cos\theta}{2g} \)

Hint:

Maximum height depends only on the vertical component of velocity, not on horizontal motion.

Answer 1

The Correct Option is A

Step 1: Resolving initial velocity.
The vertical component of initial velocity is: \[ u_y = u \sin\theta \]
Step 2: Using equation of motion.
At maximum height, final vertical velocity becomes zero. Using: \[ v^2 = u^2 - 2gh \] \[ 0 = (u\sin\theta)^2 - 2gh \]
Step 3: Solving for maximum height.
\[ h = \frac{u^2 \sin^2\theta}{2g} \]
Step 4: Conclusion.
The maximum height attained by the projectile is \[ \dfrac{u^2 \sin^2\theta}{2g} \]