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 = 22.0 m/s and at an angle of 0 = 39.0°.

(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)
Vo = 22 m/s
horizontal, Voh = 22*cos 39 = 17.1 m/s
vertical, Vov = 22*sin 39 = 13.8 m/s

AT maximum height, vertical compoenent becomes 0
Only Horizontal compoenent remains and that doesn't change

So speed will be 17.1 m/s

b)
Max height be H
then consider only vertical moton:
Vfv^2 = Vov^2 + 2*a*h
0 = 13.8^2 +2*(-9.8)*H
H=9.72 m

we need to find vertical component at h = 9.72/2 =4.86 m
Vfv^2 = Vov^2 + 2*a*h
= 13.8^2 +2*(-9.8)*4.86
Vfv=9.8 m/s
Vfh = 17.1 m/s (Horizontal compoenent doesn't change)

so speed= sqrt (9.8^2 + 17.1^2) = 19.7 m/s

c)
% = (19.7-17.1)*100 / 17.1
= 15.2 %
Answer: 15.2 %

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