Question

A trebuchet wasa hurling machine builtto attackthe 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 itfrom the castle wall. Instead, itwas positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed ofv0= 28.0 mls and at an angle ofθ0= 40.0°. What is the speed of the stoneif it hits the wall (a) just as it reaches the top of its parabolic path (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

Answer:

Explanation:

Vo = 28m/s[40o].

Xo = 28*Cos40 = 21.45m/s.

Yo = 28*sin40 = 18 m/s.

a. Y = 0, V = Xo = 21.45 m/s.

b. Y^2 = Yo^2 + 2g*h = 0.

h = -Yo^2/2g = -(18^2)/-19.6 = 16.5m, 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

Answer:

when you were answer but see its to late ;/

Explanation: