#### Question

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

#### Solution 1

Speed of the ball, *u* = 40 m/s

Maximum height, *h* = 25 m

In projectile motion, the maximum height reached by a body projected at an angle *θ*, is given by the relation:

h = `(u^2sin^2theta)/(2g)`

25 = `((40)^2sin^2theta)/(2xx9.8)`

sin^{2 }*θ* = 0.30625

sin *θ* = 0.5534

∴*θ* = sin^{–1}(0.5534) = 33.60°

Horizontal range, R = `(u^2sin 2theta)/g`

= `((40)^2xxsin2xx33.60) /9.8`

=`(1600xxsin 67.2)/9.8`

=`(1600xx0.922)/9.8 = 150.53 m`

#### Solution 2

Maximum height h_"max" = 25 m; Horizonal range, R = ?

Velocity of projection, v = `40 ms^(-1)`

We know that `h_"max" = `(v^2sin^2 theta)/2g`

or `sin^2 theta = (25xx2xx9.8)/40xx40 = 0.30625 or `sin theta = 0.5534`

`theta = sin^(-1)(0.5534) = 33.6^@`

Again `R = (v^2sin 2theta)/g = (40xx40sin 67.2^@)/9.8 `

or `R = 1600/9.8 xx 0.9219 m = 150.5 m`