Question

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

Answer 1

`70" m/s"`

Explanation:

Let the velocity of the cannonball be u m/s.The angle of projection being `alpha=45^@` the horizontal component will be `ucosalpha=ucos45=u/sqrt2` and its vertical component will be `usinalpha=usin45=u/sqrt2`

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

`0=u/sqrt2*T-1/2*9.8*T^2`

`T=u/(4.9sqrt2)`

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

`R=u/sqrt2xxT=u/sqrt2xxu/(sqrt2*4.9)=u^2/9.8`

Given `R=500m`

So `u^2/9.8=500`

`=>u^2=500xx9.8=4900`

`=>u=70" m/s"`