Q124 / 160 · AP EAMCET 2024 (21 May Shift 1) · Physics
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
- A18 m
- B36 m
- Correct answerC12 m
- D24 m
Solution
$\mathrm{R}=50 \mathrm{~m}, \theta=45^{\circ}$
By equation of trajectory $y=x \tan \theta\left(1-\frac{x}{R}\right)$
At $\mathrm{x}=20 \mathrm{~m}, \mathrm{y}=\mathrm{h}$ $\therefore \quad h=20 \tan 45^{\circ}\left(1-\frac{20}{50}\right)=20 \times \frac{3}{5}=12 \mathrm{~m}$
By equation of trajectory $y=x \tan \theta\left(1-\frac{x}{R}\right)$
At $\mathrm{x}=20 \mathrm{~m}, \mathrm{y}=\mathrm{h}$ $\therefore \quad h=20 \tan 45^{\circ}\left(1-\frac{20}{50}\right)=20 \times \frac{3}{5}=12 \mathrm{~m}$
Hence, the correct answer is option: 'C'.