A trebuchet was a hurling machine built to attack the walls of a castle under si

Question

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of v0 = 31.0 m/s and at an angle of 0 = 42.2°.

(a) What is the speed of the stone if it hits the wall just as it reaches the top of its parabolic path?

(b) What is the speed of the stone if it hits the wall when it has descended to half that height?

(c) As a percentage, how much faster is it moving in part (b) than in part (a)?

Explanation / Answer

a) The horizontal speed stays the same (disregarding air resistance). It equals 27m/s * cos(41deg) = 20.38m/s.

b) This requires the vertical component of v0. 27m/s * sin(41deg) = 17.71m/s. Then we need the time before the vertical component is 0m/s. vF = v0 + a*t, where vF = 0, v0 = 17.71m/s, a = -9.81m/s/s. So: 0 = 17.71 - 9.81*t. t = 17.71/9.81 = 1.81 seconds. Then the vertical distance is: d = (v0+vF)/2 * t; d = (17.71 + 0) / 2 * 1.81 = 16.03 meters. Half of this height is 8.02 meters. d = v0*t + .5*a*t^2; 8.02m = 0*t + .5*9.81*t*t; t = Sqrt(8.02/(.5*9.81)) = Sqrt(1.635) = 1.28 seconds. So to fall half of the 16.03 meters take 1.28 seconds. Vertical velocity: vF = v0 + a*t = 0 + -9.81*1.28 = -12.56 m/s vertically. So the rock is moving -12.56m/s vertically, and still moving 20.38m/s horizontally. A^2 + B^2 = C^2; so (-12.56^2) + (20.38^2) = v_1.28^2; v_1.28 = 23.94m/s total speed.

c) A is N percent of B: A/B = N/100 = 23.94/20.38 = 1.17. That's a 17% increase in overall speed.

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