# A cannon fires a cannonball 500m downrange when set at a 45 degree angle. At what velocity does the cannonball leave the cannon?

Nov 1, 2016

$70 \text{ m/s}$

#### Explanation:

Let the velocity of the cannonball be u m/s.The angle of projection being $\alpha = {45}^{\circ}$ the horizontal component will be $u \cos \alpha = u \cos 45 = \frac{u}{\sqrt{2}}$ and its vertical component will be $u \sin \alpha = u \sin 45 = \frac{u}{\sqrt{2}}$

If T be the time of flight then we can write net vertical displacement during this time will be zero.

$0 = \frac{u}{\sqrt{2}} \cdot T - \frac{1}{2} \cdot 9.8 \cdot {T}^{2}$

$T = \frac{u}{4.9 \sqrt{2}}$

The horizontal displacement during this time i.e the range will be

$R = \frac{u}{\sqrt{2}} \times T = \frac{u}{\sqrt{2}} \times \frac{u}{\sqrt{2} \cdot 4.9} = {u}^{2} / 9.8$

Given $R = 500 m$

So ${u}^{2} / 9.8 = 500$

$\implies {u}^{2} = 500 \times 9.8 = 4900$

$\implies u = 70 \text{ m/s}$