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 $H = \frac{u^2 \sin^2 \alpha}{2g}$.