# Initial Velocity question

derpingmath

## Homework Statement

A cannon ball weighing 128 pounds is fired from a cannon at 45 degrees above the horizontal. The cannon ball lands in a pool that has a radius of 14 feet and located 7.5 feet above the ground. The closest edge of the net is 98 feet from the cannon. It takes 4 seconds for the cannon ball to travel through the air and land in the exact center of the pool. What was the initial velocity of the cannon ball in ft/s?

## Homework Equations

I am completely lost on this. I think I should start at range and work backwards. I could really use help understanding how to break this down.

## The Attempt at a Solution

I don't have Vi which is what I am solving for. I have time (t)=4/s. I know d=vt and to solve for v it would be v=d/t. I think the distance traveled would be 98+14=112/ft. so v=28ft/s. I think at this point I have d,t,v. This is where I am lost. Am I even right on the distance, and where to go from here. I do not understand this.

Nessdude14
Hello, welcome to Physics Forums.

You're doing everything right in the x-dimension, but you also need to consider the y-dimension. You found that since the ball travels 112 feet in the x-direction, its velocity in the x-direction is a constant 28 m/s. I would call the point at 112 feet xf (for x final) and the velocity in the x direction vx.

Since the ball's vx is a constant 28 m/s, we can also say that its initial velocity (vix) is 28 m/s. Now, since the ball was fired at 45°, this makes the initial velocity (vi) a vector which is 45° from the x-axis. We know that the x-component of this vector is vix, and we know the angle of the vector, so:

$\displaystyle cos(45°)=\frac{v_{ix}}{v_i}$

$\displaystyle v_i=\frac{v_{ix}}{cos(45°)}$

derpingmath
still a little lost. Doesnt the 7.5 feet above the ground have something to do with it?

azizlwl
A cannon ball weighing 128 pounds is fired from a cannon at 45 degrees above the horizontal. The cannon ball lands in a pool that has a radius of 14 feet and located 7.5 feet above the ground. The closest edge of the net is 98 feet from the cannon. It takes 4 seconds for the cannon ball to travel through the air and land in the exact center of the pool. What was the initial velocity of the cannon ball in ft/s?
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May I know what the underline sentence refers to?

derpingmath
should be the closest edge of the pool.

So the pools edge is 98 feet from the cannon.

Cannon ---------------98ft---------------Pool's edge

azizlwl
I don't have Vi which is what I am solving for. I have time (t)=4/s. I know d=vt and to solve for v it would be v=d/t. I think the distance traveled would be 98+14=112/ft. so v=28ft/s. I think at this point I have d,t,v. This is where I am lost. Am I even right on the distance, and where to go from here. I do not understand this
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You have found vx.
V is the resultant of vx and vy.
The angle of projection is given , use trig. function to solve for v.

Nessdude14
still a little lost. Doesnt the 7.5 feet above the ground have something to do with it?

You don't need to use that information to solve the problem. Since you found one component of the initial velocity (vix), and the problem tells you that the initial velocity is angled at 45°, that's all the information you need. You simply need to calculate the magnitude of the vi vector as I described above.