Question

The maximum height of a projectile on a horizontal plane is: ____.

• A. u² sin² $\alpha$ 2g
• B. u² cos² $\alpha$ 2g
• C. u² sin² $\alpha$ g
• D. u² cos² $\alpha$ g

Hint:

Don't confuse the formula for Range ($\sin 2\alpha$) with the formula for Height ($\sin^2 \alpha$). Height depends on the square of the vertical velocity component divided by $2g$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The maximum height ($H$) is the highest vertical point reached by the projectile. At this specific point, the vertical component of the velocity becomes momentarily zero.

Step 2: Key Formula or Approach:
Using the third equation of motion for the vertical direction: \[ v_y^2 = u_y^2 - 2gH \] where $v_y = 0$ at the highest point and $u_y = u \sin \alpha$.

Step 3: Detailed Explanation:
Substitute the values into the kinematic equation: \[ 0 = (u \sin \alpha)^2 - 2gH \] \[ 2gH = u^2 \sin^2 \alpha \] Solving for $H$: \[ H = \frac{u^2 \sin^2 \alpha}{2g} \]

Step 4: Final Answer:
The maximum height is given by $H = \frac{u^2 \sin^2 \alpha}{2g}$.

Answer 2

This question asks for the maximum height reached by a projectile launched at angle \( \alpha \) with speed \( u \). Let's check each option using the physical requirement that maximum height should depend only on the vertical component of the initial velocity.

1. \( \dfrac{u^2\sin^2\alpha}{2g} \): This depends on \( \sin^2\alpha \), the square of the vertical velocity component (\( u\sin\alpha \)) divided by \( u^2 \). It correctly predicts zero height for a purely horizontal launch (\( \alpha = 0 \)) and the greatest possible height for a straight-up vertical launch (\( \alpha = 90^\circ \)), both agreeing with physical intuition.
2. \( \dfrac{u^2\cos^2\alpha}{2g} \): Using \( \cos^2\alpha \) instead would predict maximum height at \( \alpha = 0^\circ \), meaning a horizontal launch reaches the greatest height, which contradicts the fact that a horizontal launch never rises above its starting level at all.
3. \( \dfrac{u^2\sin^2\alpha}{g} \): This has the right angular dependence but is missing the factor of 2 in the denominator that comes from the underlying kinematic relationship, making it twice as large as the correct height.
4. \( \dfrac{u^2\cos^2\alpha}{g} \): This combines both the wrong trigonometric dependence and the missing factor of 2 at once, compounding both errors together.

Only the expression using \( \sin^2\alpha \) with a denominator of \( 2g \) correctly matches both the expected angular dependence and the numerical factor from the underlying kinematics.

Therefore, the correct answer is \( \dfrac{u^2\sin^2\alpha}{2g} \).