Question

A trebuchet was a hurting machine built to attack the walls of a caste 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 with a speed of v_s = 40.0 m/s and at an angle of 44.8degree. What is the speed of the stone if it hits the wall just as it reaches the top of its parabolic path? What is the speed of the stone if it hits the wall when it has descended to half that height? As a percentage, how much faster is it moving in part than in part (a)?

Answer 1

Solution: Given that Initial velocity vo 40 m/s angle θ 44. a) At the maximum point, the vertical velocity component becomes zero. But the horizontal component is non-zero Hence the velocity at that pointis v-v,-vo cos (44) = 28.77 m/s (b)The maximum height of the object is u2 _ (voxsin44). 39.39 m Hence the half of H is h-19.695 m the x-component of the velocity is same in all situation Vx-Vo cos (44) 28.77 m Now the y-component of th velocity at h is v, -V sin (44)) -2x9.8x 19.695 v, -19.648 m/s ce vvj+v v-34.839 m/s V28.77+19.648

(c) Diffence in velocity is Av-34.839-28-77-6.069 m/s Hence percentage change is 6.069×100= 21.0 Δνο o_ww77 100-21.09% 28.77