Question

The ceiling of a long hall is 25 $\text m$ high. What is the maximum horizontal distance that a ball thrown with a speed of 40$\text m \;\text s^{-1}$ can go without hitting the ceiling of the hall ?

Answer 1

Speed of the ball, $\text u$ = 40 $\text m/\text s$
Maximum height, $\text h$ = 25 $\text m$
In projectile motion, the maximum height reached by a body projected at an angle $\theta$, is given by the relation:

$\text h$ = $\frac{\text u^2 \text {sin}^2\theta}{2\text g}$

$25$ = $\frac{(40)^2 \sin^2 \theta}{2\times 9.8}$

$\sin^2 \theta$ = 0.30625
$\sin \theta$ = 0.5534
∴ $\theta$ = $\sin^{-1}$ (0.5534) = 33.60°

Horizontal range, $\text R$ = $\frac{\text u^2 \sin2 \theta}{\text g}$

= $\frac{(40)^2 \times \sin2\times 33.60}{ 9.8}$

= $\frac{1600 \times \sin 67.2}{9.8}$

= $\frac{1600 \times 0.992 }{9.8 }$= $150.53 \;\text m$