Question

The horizontal range of a projectile projected at an angle of \( 45^\circ \) with the horizontal is 50 m. The height of the projectile when its horizontal displacement is 20 m is:

• A. \( 18 \) m
• B. \( 36 \) m
• C. \( 12 \) m
• D. \( 24 \) m

Hint:

For projectile motion, use the equation \( y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \) to determine height at a given horizontal displacement.

Answer 1

The Correct Option is C

Step 1: Given parameters
- Angle of projection: \( 45^\circ \).
- Horizontal range: \( R = 50 \) m.
- Horizontal displacement: \( x = 20 \) m.
The equation for the trajectory of a projectile is: \[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}. \] For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \) and \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), so the equation simplifies to: \[ y = x - \frac{g x^2}{2 u^2 \cos^2 45^\circ}. \] Since the horizontal range is given by: \[ R = \frac{u^2 \sin 2\theta}{g}, \] for \( \theta = 45^\circ \), we get: \[ 50 = \frac{u^2}{g}. \] Thus, \[ u^2 = 50g. \] Step 2: Calculating the height
Substituting \( u^2 = 50g \) and \( \cos^2 45^\circ = \frac{1}{2} \): \[ y = x - \frac{g x^2}{2 (50g) \times \frac{1}{2}}. \] \[ y = x - \frac{x^2}{50}. \] Substituting \( x = 20 \): \[ y = 20 - \frac{20^2}{50}. \] \[ y = 20 - \frac{400}{50}. \] \[ y = 20 - 8 = 12 { m}. \] Step 3: Conclusion
Thus, the height of the projectile when the horizontal displacement is 20 m is: \[ 12 { m}. \]