Question

A ball is thrown at an angle of \({\rm{4}}{{\rm{5}}^{\rm{0}}}\)to the ground. If the ball lands 90 m away, what was the initial speed of the ball?

Short answer

The initial speed of the ball is \(30m/s\).

Step-by-step solution

First equation of motion

The first equation of motion is\({\rm{v = u + at}}\); where u and v are the initial and final velocity. a is the acceleration and t is the time period.

Finding initial speed of the ball

Let the initial speed of the ball be u.

Initial upward speed can be written as-

\(\begin{aligned}{c}u\sin \theta = u\sin {45^0}\\ = \frac{u}{{\sqrt 2 }}\end{aligned}\)

i.e., the initial vertical velocity is \(\frac{u}{{\sqrt 2 }}\) and final vertical velocity is \(v = - \frac{u}{{\sqrt 2 }}\).

Vertical acceleration will be constant which is \( - 10m/{s^2}\).

Using first equation of motion,

\(\begin{aligned}{c}v = u + at\\t = \frac{{v - u}}{a}\end{aligned}\)

\(\begin{aligned}{c}t = \frac{{ - \frac{u}{{\sqrt 2 }} - \frac{u}{{\sqrt 2 }}}}{{ - \sqrt 2 (10)}}\\t = \frac{u}{{5\sqrt 2 }}\end{aligned}\)

Distance traveled by the ball can be written as-

\({\rm{distance = velocity \times time}}\)

\(\begin{aligned}{c}{\rm{90 = }}\frac{u}{{5\sqrt 2 }} \times u\cos \theta \\ = \frac{u}{{5\sqrt 2 }} \times \frac{u}{{\sqrt 2 }}\\ = \frac{{{u^2}}}{{10}}\\{u^2} = 900\\u = 30\end{aligned}\)

Therefore, the initial speed of the ball is \(30m/s\).