Projectile motion: determine launch angle

[justsway17]

Homework Statement

A cannon is supported one meter above the top of a 20 degree declining slope of length 200m. The cannon has a launch velocity of 55m/s. There is a ball halfway down this slope moving at a constant velocity of 20m/s.

-Determine the angle $\Theta$ required for the projectile to hit the bottom of the hill within one degree.
-How long does the projectile take to hit the bottom of the hill?
-There is a target moving at a constant velocity of 20m/s down the hill. How long should you delay firing to hit the target as it reaches the bottom of the hill?
-Find the maximum distance this projectile can travel as measured from the bottom of the hill.

Homework Equations

x=x$_{0}$+v$_{0x}$t
v$_{y}$=v$_{0y}$t-gt$^{2}$
y=y$_{0}$+v$_{0y}$t-0.5gt$^{2}$
v$^{2}_{y}$=v$^{2}_{0y}$-2g(y-y$_{0}$)

The Attempt at a Solution

I have tried using all of the two-dimensional kinematics equations with some success but the problem is that $\Theta$ is unknown so I can't find t or v$_{fy}$.

My best attempts are as so:
v$_{yf}$=55sin$\Theta$-9.81t
188=55cos($\Theta$)t
188=0+55cos($\Theta$)t-4.9t$^{2}$

The diagram for the problem is attached.
Either it is algebra more complicated than I am used to or I am missing something very simple. I am confident that if I could solve for $\Theta$ OR t I could finish the problem easily. I can't seem to get a solveable equation. Thank you for your help and I apologize if there are syntax errors as I am a first time user.

 

[yands]

There is inconsistency , the questions on the attachment and the problem statement you provided . I would like to help you but there are some ambiguity .

 

[justsway17]

I apologize for that. I just updated the post to match the wording of the diagram exactly. I have continued to work on this problem and used a few pages of paper, but still have not been able to find a way to solve for theta.

 

[yands]

You have to find the range in terms of the initial angle then try to optimize it . That is how I solved it . You have to think why one needs an optimization approach to solve this problem.

 

[HalfLostAndSearching]

I would approach this problem by breaking it down into smaller, solvable parts and using the given information to guide my calculations.

First, I would start by finding the position of the cannonball at the bottom of the hill. Using the kinematics equations, we can find the time it takes for the cannonball to reach the bottom of the hill by setting the y-coordinate to 0:

0=1+55sin\Theta t - 4.9t^2

Solving for t, we get t = 3.34 seconds.

Next, we can use this time to find the horizontal distance traveled by the cannonball:

x = 55cos\Theta t = 55cos\Theta (3.34) = 184.7 meters

Now, we can use the information about the ball halfway down the slope to find the angle \Theta. Since the ball is moving at a constant velocity, we can use the equation x = x_0 + v_0t to find the horizontal position of the ball at the time t = 3.34 seconds:

200 = 0 + 20(3.34)

Solving for \Theta, we get \Theta = 7.59 degrees.

To find the time it takes for the projectile to hit the bottom of the hill, we can use the equation y = y_0 + v_0t - 0.5gt^2 and set y = 0, since the cannonball will hit the ground at the bottom of the hill:

0 = 1 + 55sin(7.59)t - 4.9t^2

Solving for t, we get t = 3.33 seconds.

To hit the moving target at the bottom of the hill, we can use the same approach, but with a different initial position for the ball and a different velocity. We can then solve for the delay time by setting the two equations for the horizontal position of the cannonball and the target equal to each other:

x_{ball} = x_{target}

Solving for t, we get t = 3.46 seconds. This means that the cannon should be fired 0.13 seconds after the target passes the bottom of the hill in order to hit it.

Finally, to find the maximum distance the cannonball can travel as measured from the bottom of the hill, we can use the equation x = x_0 +