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, \(\text R\) = \(\frac{\text u^2 \sin2 \theta}{\text g}\) 

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

= \(\frac{1600 \times \sin 67.2}{9.8}\)

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