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 = 25.0 m/s and at an angle of ?0 = 46.9°.
What is the speed of the stone if it hits the wall just as it reaches the top of its parabolic path?

Explanation / Answer

For trajectories, the initial velocity has x and y components as: Vx = Vcos(T) ; horizontal component Vy = Vsin(T) ; vertical component You can derive the equations below from the basic equations of motion. Max height: h = Vy²/2g h = v²sin²(T)/2g Max distance: d = v²sin(2T)/g ; trig identity sin(2T) = 2sin(T)cos(T) d = v²2sin(T)cos(T)/g Total time in flight: t = 2 Vy/g t = 2 Vsin(T)/g For (a) the speed would be the initial horizontal velocity only; the vertical velocity would be 0. For (b) the speed would be a combination of the initial horizontal velocity and the vertical velocity achieved dropping half the max height. For (c) calculate the ratio of (b) to (a).

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