Question

# A cricketer can throw a ball to a maximum horizontal distance of $100\text{m}$. How much high above the ground can the cricketer throw the same ball?

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Solution

## Step 1: GivenThe distance the cricketer can throw the ball horizontally, $R=100\text{m}$Step 2: Formulas usedFor a projectile motion, we know that the range (horizontal distance covered) is given as,$R=\frac{{V}^{2}\mathrm{sin}\left(2\theta \right)}{g}$where $V$ is the initial velocity, $\theta$ is the angle at which object was thrown at and $g$ is the acceleration due to gravity.And the maximum height reached is given as,$H=\frac{{V}^{2}{\mathrm{sin}}^{2}\left(\theta \right)}{2g}$where the symbols mean the same as described earlier.Step 3: Calculating heightWe know that the maximum horizontal distance is covered when $\theta =45°$.So we have,$\begin{array}{ccc}& R=\frac{{V}^{2}\mathrm{sin}\left(2\theta \right)}{g}& \\ ⇒& 100=\frac{{V}^{2}\mathrm{sin}\left(2×\frac{\mathrm{\pi }}{4}\right)}{g}& \left[\because 45°=\frac{\mathrm{\pi }}{4}\text{rad}\right]\\ ⇒& \frac{{V}^{2}}{g}=100& \left[\because \mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)=1\right]\dots \left(1\right)\end{array}$We also have that,$\begin{array}{ccc}& H=\frac{{V}^{2}{\mathrm{sin}}^{2}\left(\theta \right)}{2g}& \\ ⇒& H=\frac{100{\mathrm{sin}}^{2}\left(\frac{\mathrm{\pi }}{4}\right)}{2}& \left[\because \left(1\right)\right]\\ ⇒& H=50{\left(\frac{1}{\sqrt{2}}\right)}^{2}& \left[\because \mathrm{sin}\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}}\right]\\ ⇒& H=25\text{m}& \end{array}$Therefore, the maximum height the cricketer can throw the ball is $25\text{m}$.

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