Question

What is the relation between the height $h$ and the range $R$ when a projectile is fired at $45^\circ$?

• A. $R = h$
• B. $R = 2h$
• C. $R = 4h$
• D. $R = 8h$

Hint:

For a projectile at $45^\circ$: Range = $4 \times$ Maximum Height.

Answer 1

The Correct Option is C

Concept:
In projectile motion, the maximum height and horizontal range depend on the initial velocity and the angle of projection. When the angle of projection is $45^\circ$, the projectile achieves maximum horizontal range.

Step 1: Formula for maximum height \[ h = \frac{u^2 \sin^2\theta}{2g} \] For $\theta = 45^\circ$: \[ \sin^2 45^\circ = \frac{1}{2} \] \[ h = \frac{u^2}{4g} \]

Step 2: Formula for horizontal range \[ R = \frac{u^2 \sin 2\theta}{g} \] For $\theta = 45^\circ$: \[ \sin 90^\circ = 1 \] \[ R = \frac{u^2}{g} \]

Step 3: Find relation between $R$ and $h$ \[ h = \frac{u^2}{4g} \] \[ R = \frac{u^2}{g} \] \[ R = 4h \] Conclusion:
Thus, when a projectile is fired at $45^\circ$, the relation between range and maximum height is $\mathbf{R = 4h}$.