Can the Catapult Clear the Wall Without Violating Safety Rules?

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Homework Statement

You and your friend have been hired to see if the catapult a movie company owns will be usable for a scene about a battle set in the middle ages. In the scene a catapult launches a 750 kg boulder that is supposed to hit the ground 350 m away. The director is worried the boulder will hit a 7.0 m high stone wall the movie company built 325 m from the catapult. You have measured the angle the boulder leaves the catapult relative to the ground to be 30 degrees. Because there is an airport near the movie filming location, there is a safety rule that the boulder cannot fly over 200 m high. Assume that the catapult is 2 m off the ground and the movie set is on flat ground. Can the catapult launch the stone and not violate the safety rule so that the stone clears the wall? If not, how high on the wall will it hit and if so, by how much will the boulder miss the top of the wall?

Homework Equations

d=vit+1/2at²
v2² = v1² +2ad
vix=d/t
x=(vo²sin2theta)/g
Fnet=ma

The Attempt at a Solution

My main issue with this question is the fact that the catapult is launched from 2 m above the ground but it lands on the ground. I tried using the 2 m point as the "ground" or reference line, but that didn't really help because it messes with the range (the projectile travels 350 m when it is launched from 2 m above the ground, not when it is launched from the ground. If it is launched from the ground, the range wouldn't be 350 m). This makes it difficult to find the maximum height (which must be less than 200 m to meet the safety guideline). I tried using the equation horizontal range=[(vo)²(sin2theta)]/g, but that only works if the object returns to the level it was launched at, and in this case it doesn't, so I don't know how else to find the inital velocity only being given the mass of the boulder and the angle it is launched at. I tried splitting the diagram into two halves and seeing whether I could examine the motion of the boulder during the second half (with a range of 175 m, starting from the highest point where vy=0) but this didn't work because I did not know the initial velocity. Thanks in advance for your help! :)

 

[rl.bhat]

The general expression for a projectile is given by
y = tan(theta)*x - g*x^2/(2Vo^2*cos^2theta).
In the first case take y = -2m because it is along the negative y axis, and in the second case take y as ( 7m - 2m ) = 5m. Solve the two equations to find Vo