Finding initial velocity of a cannonball fired at 45 degrees over 100 meters

[MisterNi]

Ok, here's the problem:

A cannon angled at 45 degrees above the horizontal is fired. The cannonball lands 100 meters away. What is the initial velocity?

I've tried using the kinematic equations, but I always end up with at least 2 variables with 1 equation. I'm just completely stumped. Any help to nudge me in the right direction would be greatly appreciated. Thank you.

 

[Maxos]

Post your calculations so that we can show you the solution/errors.

 

[BobG]

You have two equations: one to solve for the horizontal displacement and one to solve for the vertical displacement.

 

[HallsofIvy]

You certainly should have two equations- one for the horizontal position and one for the vertical position. Your two variables will be the horizontal component of initial velocity (which is equal to the vertical component because the angle is 45 degrees) and time.

 

[mukundpa]

May use the formula for horizontal range of a projectile

R = u^2sin(2theeta)/g

 

[Danger]

I'd say that your army is in trouble if you can only shoot 100 metres.
Actually, I just wanted to clarify if this is neglecting air resistance. That makes a big difference.

 

[MisterNi]

Thanks for BobG, HallsofIvy and mukundpa, I think I got the answer.