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: The vertical component of the initial velocity is \( u_y = u\sin\theta \), while gravity decelerates only this vertical component during the rise.
Step 2: At the maximum height, the vertical velocity becomes zero, so using \( v^2 = u_y^2 - 2gh \) with \( v=0 \) gives \( 0 = (u\sin\theta)^2 - 2gh \).
Step 3: Rearranging for \( h \). \[ \boxed{h = \dfrac{u^2\sin^2\theta}{2g}} \]

Answer 2

Another way to reach the same result is by checking units and limiting cases for each option rather than deriving the formula from scratch.

1. Option \( \dfrac{u^2\sin^2\theta}{2g} \): This has units of (velocity)\(^2\)/(acceleration), which works out to length, matching what a height should be; additionally, checking the limiting case \( \theta = 90^\circ \) (straight up), this reduces to \( \dfrac{u^2}{2g} \), the well-known maximum height for a purely vertical throw, confirming the formula behaves correctly at this extreme.
2. Option \( \dfrac{u^2\cos^2\theta}{2g} \): At the same limiting case \( \theta = 90^\circ \), \( \cos(90^\circ) = 0 \), so this expression would predict zero maximum height for a projectile thrown straight up, which is clearly wrong since a vertically thrown projectile does rise to a height of \( u^2/2g \); this contradiction rules the option out.
3. Option \( \dfrac{u\sin\theta}{2g} \): Checking units, velocity divided by acceleration gives units of time, not length, so this cannot represent a height at all, independent of any limiting-case check.
4. Option \( \dfrac{u\cos\theta}{2g} \): This shares the same unit mismatch (time instead of length) as the previous option, and additionally fails the \( \theta=90^\circ \) check by predicting zero height for a straight-up throw.

Checking both dimensional consistency and the straight-up throw limiting case rules out three of the four options immediately.

So the correct answer is \( \dfrac{u^2\sin^2\theta}{2g} \).