Cannonball Projectile Motion

[mattwild]

Homework Statement

A cannon has a launch speed of 500m/s and is aimed at a target 2.8km away.
a. Write an equation for the maximum horizontal range of the cannon as a function of the launch angle theta.
b. At what angle must the cannon be launched in order to hit the target?
c. As viewed from above, how quickly does the cannonball's shadow move along the ground?
d. How fast is the cannonball moving when it hits its target?

Homework Equations

D = vit+1/2at^2
velocity formulas

Range of a projectile

The Attempt at a Solution

a. I used the range of a projectile motion and got v^2sin2(theta)/g but I don't know how to make it a function of the launch angle theta. Am I overreading?
b. I am assuming you just plug in the values from the first formula you get.
c. Just find the acceleration?
d. Just find the vf?

 

[mfb]

a. I used the range of a projectile motion and got v^2sin2(theta)/g but I don't know how to make it a function of the launch angle theta. Am I overreading?

d= v^2sin2(theta)/g is a formula for the range as function of the launch angle (and constants). I used d= with d as range to make a formula out of your expression.

b. I am assuming you just plug in the values from the first formula you get.

Right.

c. Just find the acceleration?

Why acceleration?

d. Just find the vf?

There is an easier way, but that works as well.

 

[mattwild]

Because it asks how quickly does the cannonball's shadow move along the ground so I am assuming that means acceleration? Because its velocity is constantly changing so it can't be that

 

[mfb]

Because it asks how quickly does the cannonball's shadow move along the ground so I am assuming that means acceleration?

No, that certainly means velocity. The velocity of the shadow is not changing here if the sun is shining vertically.

 

[Nickie]

a. Yes, you are correct in using the range formula for projectile motion. To make it a function of the launch angle theta, you can simply replace the angle in the formula with theta. So your equation would be: R = (v^2sin2(theta))/g, where R is the range and v is the launch speed.

b. To find the angle at which the cannon must be launched to hit the target, you can rearrange the formula to solve for theta. So your equation would be: theta = (1/2)arcsin(gR/v^2). Plugging in the values of g, R, and v, you can calculate the angle theta.

c. To find how quickly the cannonball's shadow moves along the ground, you can use the velocity formula for projectile motion, which is v = v0 + at. Since the cannonball's initial velocity is 500m/s and the acceleration due to gravity is 9.8m/s^2, the velocity of the cannonball at any given time will be 500m/s - 9.8m/s^2t. The shadow will move at the same velocity as the cannonball, so the speed of the shadow can be calculated using the same formula.

d. To find the speed of the cannonball when it hits its target, you can use the velocity formula again. This time, you will be solving for the final velocity, vf. So your equation would be: vf = v0 + at. Plugging in the values of v0 (500m/s) and a (9.8m/s^2), you can calculate the final velocity vf.