Question

1. During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +40.0 m/s at an angle of 55.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.

Answer 1

Initial velocity u = 40 
Angle at launch = 55 degrees
 At maximum height v = 0, velocity equation v^2 = u^2 - 2gh,
 0 = (40 x sin55)^2 - 2 x 9.81 x h => (40 x 0.819)^2 = 19.62h => h = 32.76^2 /
19.62
 Maximum height = 54.7 m 
 We have v = u - gt => 0 = (40 x sin55) - 9.81 x t => t = 32.76 / 9.81 => t =
3.34 s
 Time taken to hit the ground is 2t = 2 x 3.34 = 6.68 s 
Distance from castle to trebuchet = utcos55 = 40 x 6.68 x 0.573 = 153.1 m