Question

A projectile shot at an angle of ${45}^o$ above the horizontal strikes the wall of a building $30m$ away at a point $15m$ above the point of projection. Initial velocity of the projectile is
(take $g=9.8m/s^2$)

• A. $14m/s$
• B. $14\sqrt{2}m/s$
• C. $14\sqrt{3}m/s$
• D. $14\sqrt{5}m/s$

Answer 1

Answer: (C)

$14\sqrt{3}m/s$

Equation of trajectory:

$y=xtan\theta -\frac{g{x}^2}{2u^2{cos}^2\theta }$

The point $(30,15)$ satisfies this equation.

$15=30tan{45}^0-\frac{9.8\times {30}^2}{2u^2{cos}^2{45}^0}\Rightarrow u^2=\frac{8820}{15}=588\Rightarrow u=14\sqrt{3}$