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

Step 1: - Calculate the value of $\theta$ (angle made with horizontal).
As per question, a ball is thrown in a hall such that it should not hit the ceiling of the hall and given, the speed of the ball $u=40m{s}^{-1}$.
As we know that, in projectile motion, the maximum height reached by a body projected at an angle θ, is given by,
$h_{max}=\frac{u^2{\sin }^2\theta }{2g}$
Substituting the values, we get,
$25=\frac{(40{)}^2{\sin }^2\theta }{2\times 9.8}$
${\sin }^2\theta =\frac{25\times 2\times 9.8}{1600}=0.30625$
$\sin \theta =\sqrt{0.30625}=0.5534$
$\Rightarrow \theta ={\sin }^{-1}\left(0.5534\right)={33.60}^{\circ }$

Step 2: - Find horizontal range in projectile motion.
As the horizontal range is given by
$R=\frac{u^2\sin 2\theta }{g}$
Substituting the value of u, g and $\theta$, we get
$R=\frac{\left(40{)}^2\sin \right(2\times {33.60}^{\circ })}{9.8}$
$R=\frac{1600\times 0.922}{9.8}=150.53m$

Final Answer: The maximum horizontal distance that a ball thrown, can go without hitting the ceiling of the hall is 150.53 m.