The horizontal range of a projectile projected at an angle of $45^{\circ}$ with the horizontal is 50 m . The…

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
  1. 18 m
  2. 36 m
  3. 12 m
  4. 24 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 $\mathrm{x}=20 \mathrm{~m}, \mathrm{y}=\mathrm{h}$ $\therefore \quad h=20 \tan 45^{\circ}\left(1-\frac{20}{50}\right)=20 \times \frac{3}{5}=12 \mathrm{~m}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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