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: At launch, the projectile has total kinetic energy \( \dfrac{1}{2}mu^2 \). At the highest point, the vertical velocity is zero and only the horizontal component \( u\cos\alpha \) remains, so the kinetic energy there is \( \dfrac{1}{2}m(u\cos\alpha)^2 \), plus the gained potential energy \( mgH \).

Step 2: Apply conservation of energy between launch and the highest point: \( \dfrac{1}{2}mu^2 = \dfrac{1}{2}m(u\cos\alpha)^2 + mgH \).

Step 3: Cancel the mass \( m \) and solve for \( H \): \( gH = \dfrac{1}{2}u^2(1 - \cos^2\alpha) = \dfrac{1}{2}u^2\sin^2\alpha \), using the identity \( 1 - \cos^2\alpha = \sin^2\alpha \).
\[ \boxed{H = \dfrac{u^2\sin^2\alpha}{2g}} \]