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: \[ 12 { m}. \]

Answer 2

To find the height of the projectile when its horizontal displacement is 20 m, we begin by understanding the path of a projectile. A projectile launched at angle \( \theta = 45^\circ \) with an initial velocity \( u \) achieves its range using the formula:

\[ R = \frac{u^2 \sin 2\theta}{g} \]

Given \( R = 50 \) m and \( \theta = 45^\circ \), the formula becomes:

\[ 50 = \frac{u^2 \sin 90^\circ}{g} \]

Since \( \sin 90^\circ = 1 \), it simplifies to:

\[ 50 = \frac{u^2}{g} \]

This implies:

\[ u^2 = 50g \]

Now, to find the height when horizontal displacement (x) is 20 m, the horizontal displacement equation is:

\[ x = u \cos\theta \cdot t \]

With \( \theta = 45^\circ \), \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):

\[ 20 = \frac{u}{\sqrt{2}} \cdot t \]

Solving for \( t \), we get:

\[ t = \frac{20 \sqrt{2}}{u} \]

The vertical displacement \( y \) is given by:

\[ y = u \sin\theta \cdot t - \frac{1}{2} g t^2 \]

Substitute \( \sin 45^\circ = \frac{1}{\sqrt{2}} \) and \( t = \frac{20 \sqrt{2}}{u} \):

\[ y = \frac{u}{\sqrt{2}} \cdot \frac{20 \sqrt{2}}{u} - \frac{1}{2} g \left(\frac{20 \sqrt{2}}{u}\right)^2 \]

This simplifies to:

\[ y = 20 - \frac{1}{2} g \cdot \frac{800}{u^2} \]

Knowing \( u^2 = 50g \), the equation becomes:

\[ y = 20 - \frac{1}{2} \cdot \frac{800}{50} \]

Calculate:

\[ y = 20 - \frac{1}{2} \cdot 16 \]

\[ y = 20 - 8 \]

\[ y = 12 \text{ m} \]

Thus, the height of the projectile when its horizontal displacement is 20 m is \( 12 \) m.