You need some analysis. Break the initial velocity out of the cannon into horizontal and verticalcomponents.
Vx (horizontal)=V*cos26
Vy (vertical)=V*sin36
Now maximimum height depends on gravity, and the initial vertical velocity
At the top, at max height, the vertical velocity is zero. Find when that happens
vf=Vi+gt
0=Vsin36-9.8m/s^2 * t
solve for t. Now you know when, so
h= Vsin36*t-1/2 9.8 t^2 solve fror how high at that time.
I have a sheet full of equations, I just don't know which one to use or what type of problem this is.
A cannon is on a hill. The hill is 20 m high. The cannon is at an angle of 26 degress. If a cannonball is shot into the valley below.
a.) Find the x and y components of velocity?
so Vox=? Voy=?
b.) What is the maximum height the cannonball can reach above the valley floor?
3 answers
These are problems in ballistics. Nowadays they call them kinematics problems.
Vox is the horizintal component of the the velcoity. In order to comupte it, you need to know the speed that the cannonball leaves the cannon.
We will be glad to critique your work, but, if you are clueless, you should complain to the organization that is supposed to be teaching you.
Vox is the horizintal component of the the velcoity. In order to comupte it, you need to know the speed that the cannonball leaves the cannon.
We will be glad to critique your work, but, if you are clueless, you should complain to the organization that is supposed to be teaching you.
Where did you get the 36? Isn't it 26 in the problem or do you change it?
I would complain, but it's preparation for my midterm. I just can't remember stuff from the start of school. thanks =)
I would complain, but it's preparation for my midterm. I just can't remember stuff from the start of school. thanks =)