Question

The maximum height reached by the projectile is $4m$. The horizontal range is $12m$. The velocity of projection in $m{s}^{-1}$ Is (g - acceleration due to gravity)

Answer 1

Step 1: Given data

Maximum height $=4m$

Horizontal range $=12m$

Step 2: Formula used

Maximum height $=\frac{{v_0}^{2}si{n}^2\theta }{2g}$

Horizontal range $=\frac{{v_0}^2\sin 2\theta }{2g}$

Step 3: Find the angle of projection

It is given that the maximum height of the projectile$=4m$, thus

$4=\frac{{v_0}^{2}si{n}^2\theta }{2g}--\left(1\right)$

It is also given that the horizontal range of the projectile $=12m$ thus

$12=\frac{{v_0}^2\sin 2\theta }{2g}--\left(2\right)$

Divide equation (2) by equation (1)

$$\begin{gathered} \Rightarrow \frac{12}{4}=\frac{\sin 2\theta }{{\sin }^2\theta }\times 2 \\ \Rightarrow 3=\frac{2\sin \theta \cos \theta }{{\sin }^2\theta }\times 2 \\ \Rightarrow \frac{3}{4}=\frac{\cos \theta }{\sin \theta } \\ \Rightarrow \frac{4}{3}=\tan \theta \\ \Rightarrow \frac{4}{5}=\sin \theta \end{gathered}$$

Step 4: Find the velocity

Now, put the value of $\sin \theta$ in equation (1)

$$\begin{gathered} \Rightarrow 4=\frac{{v^2}_0{\left({\displaystyle \frac{4}{5}}\right)}^2}{2g} \\ \Rightarrow \frac{4\times 2g}{{\displaystyle \frac{16}{25}}}={v^2}_0 \\ \Rightarrow \frac{8g\times 25}{16}={v^2}_0 \\ \Rightarrow \frac{25g}{2}={v^2}_0 \\ \Rightarrow 5\sqrt{\frac{g}{2}}=v_0 \end{gathered}$$

Hence the velocity of projection is $5\sqrt{\frac{g}{2}}$.