# Cannonball is catapulted

1. Homework Statement :A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 34 m/s at an angle of 55° with respect to the horizontal and the cannonball is 8.0 m above the ground at this time.
(a) What is the maximum height above the ground reached by the cannonball?

(c) What are the x- and y-components of the cannonball's velocity just before it lands? The y-axis points up.

## Homework Equations

Maximum height = Yf=Yo+Voy*t+1/2gt^2
Voy=34sin55=27.85 m/s
Vf=Vo+at=0=27.85-9.8t t=2.8s
Maximum height= 47.56 m
How do I find the x and y components? what equations do I need?

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vela
Staff Emeritus
Homework Helper
You only have a handful of equations for projectile motion. What other ones do you have?

BruceW
Homework Helper
Are we on to part (c) now? Well, you've already got the equation for y as a function of t. From this you can get t. Then you can use t in the velocity equations.

There are no forces in the x-direction, so what will the equation for x component of velocity be? And for y-direction, there is constant force, which you already have the equation for.

Could you explain why my maximum height is wrong please?

I dont know that equations for part see...

BruceW
Homework Helper
Oh, yeah. Your max height is wrong, I didn't see that.

Well, your time for maximum height (2.8s) is correct. So you just need to put this into your equation for the height to get the maximum height.

vela
Staff Emeritus