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 \[ \dfrac{u^2 \sin^2\theta}{2g} \]

Answer 2

A projectile is launched with initial speed \( u \) at angle \( \theta \) above the ground. Its maximum height can be checked using energy conservation instead of the equations of motion, by comparing the vertical kinetic energy lost as the projectile rises to its peak against the potential energy gained.

1. Option \( \dfrac{u^2\sin^2\theta}{2g} \): At launch, the vertical component of velocity is \( u\sin\theta \), contributing vertical kinetic energy \( \tfrac{1}{2}m(u\sin\theta)^2 \) per unit consideration; at the highest point, this vertical velocity component drops to zero (only the horizontal component \( u\cos\theta \) remains), so all of the vertical kinetic energy converts into gravitational potential energy \( mgh \). Setting \( mgh = \tfrac{1}{2}m(u\sin\theta)^2 \) and solving gives \( h = \dfrac{u^2\sin^2\theta}{2g} \), matching this option exactly.
2. Option \( \dfrac{u^2\cos^2\theta}{2g} \): This would follow if the horizontal component of velocity, rather than the vertical component, were the one converting to potential energy at the peak; but the horizontal component \( u\cos\theta \) is unaffected by gravity and persists unchanged throughout the flight, so it cannot be the component responsible for the height gain.
3. Option \( \dfrac{u\sin\theta}{2g} \): This expression is not even dimensionally consistent with a height (it has units of time, not length, since velocity divided by acceleration gives a time, not a distance), so it cannot represent the maximum height regardless of the numerical reasoning.
4. Option \( \dfrac{u\cos\theta}{2g} \): This has the same dimensional inconsistency as the previous option (velocity over acceleration gives units of time, not length), ruling it out on dimensional grounds alone, in addition to using the wrong (horizontal) velocity component.

Converting the vertical kinetic energy lost while rising into potential energy gained, and checking the dimensions of each candidate expression, isolates the physically and dimensionally consistent option.

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