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

1 Answer
Nov 1, 2016

Answer:

#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"#