Solving Catapult Problem: Launch Speed, Time in Air, and Impact Speed

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SUMMARY

The discussion centers on solving a physics problem involving a catapult launching spaetzle projectiles at a Roman fortification. The key parameters include a launch height of 3.30 m, a wall height of 8.50 m, a launch angle of 60.0 degrees, and a horizontal distance of 39.1 m. Participants emphasize the importance of separating horizontal and vertical components of motion and correctly calculating the difference in heights, which is 5.2 m. The problem-solving approach involves using kinematic equations to determine launch speed, time in the air, and impact speed.

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Homework Statement


Catapults date from thousands of years ago, and were used historically to launch everything from stones to horses. During a battle in what is now Bavaria, inventive artillerymen from the united German clans launched giant spaetzle from their catapults toward a Roman fortification whose walls were 8.50 m high. The catapults launched spaetzle projectiles from a height of 3.30 m above the ground, and a distance of 39.1 m from the walls of the fortification at an angle of 60.0 degrees above the horizontal (see figure). The projectiles were to hit the top of the wall, splattering the Roman soldier atop the wall with pulverized pasta. (For the following questions, ignore any effects due to air resistance.)

What launch speed was necessary?

How long were the spaetzle in the air?

At what speed did the projectiles hit the wall?

Homework Equations



Velocity in M/S
Time in seconds
x = x(initial) + velocity*t(s)+.5*a(t^2)
Vf^2 = VI^2 + 2*a*x (x = distance)

The Attempt at a Solution



The problem is submitted via webassign and I've tried solving the right triangle made by the distance and the difference in heights of the starting location and ending location. (39.44 m) and then attempting to treat the projection with that being the total x-change and solving another right triangle to change the acceleration.

I'm rather lost I know the physics of the problem (x-velocity independent of acceleration) and y-component dependent upon the velocity up. Any tips how to start this problem would be greatly appreciated.
 
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difference in heights of the starting location and ending location. (39.44 m) and then attempting to treat the projection with that being the total x-change
Surely the difference in heights is 5.2 m: launched from 3.3 m and walls are 8.5m.

You must keep horizontal numbers separate from vertical ones all the way through. No triangles.
 
Delphi51 said:
Surely the difference in heights is 5.2 m: launched from 3.3 m and walls are 8.5m.

You must keep horizontal numbers separate from vertical ones all the way through. No triangles.

Thanks, but I was able to solve the first two questions and I am working on the last one.

I created two formulas for the summations of the x and y vectors(both of which had the total summation of the launch vector) and solved in terms of time (in air). I set equal to each other and solved.
 

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