How High Does the Cannonball Fly and What Are Its Final Velocity Components?

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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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You only have a handful of equations for projectile motion. What other ones do you have?
 
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 don't know that equations for part see...
 
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.
 
The last term in the equation for yf should have a negative sign in front of it, which you apparently used in your calculations. If you're entering the answer into the computer and it's marking it as wrong, you're probably entering it incorrectly. Perhaps you left the units off or you're using the wrong number of significant figures.

lalahelp said:
I don't know that equations for part see...
Well, what are the equations you have for projectile motion? Surely, you can find them in your book and tell us what they are. Then we can help you figure out which ones are the best to use.
 

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