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 = 27.0 m/s and at an angle of θ0 = 42.0°. What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) 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)?

Answer 1

Vo = 27m/s[42o].
Xo = 27*Cos42 = 20.1 m/s.
Yo = 27*sin42 = 18.1 m/s.

a. Y = 0, V = Xo = 20.1 m/s.

b. Y^2 = Yo^2 + 2g*h = 0.
h = -Yo^2/2g = -(18.1^2)/-19.6 = 16.7 m, max.

Y^2 = Yo^2 + 2g*h/2. = 0 + 19.6*8.35 =
163.66.
Y = 12.8 m/s.
V = Sqrt(Xo^2 + Y^2).

c. (Xo+Y) - Xo = (20.1+12.8i) - 20.1 =
12.8i = 12.8m/s[90o].

Answer 2

I don't understand why we made vy^2 to be zero first to find the height.can u please explain