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 θ₀ = 41.0°. 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

given Vo = 25 m/s

take g = 9.81 m/s^2 ( acceleration due to gravity )

Take apart the horizontal and vertical components of the velocity

the horizontal component will be constant because there is no air resistance and no horizontal forces acting on it

but the vertical component changes due to gravity acting downwards

so now

Vox = Vo * cos

= (25) * cos ( )

= 25 * 0.7547

= 18.86773 m/s

Voy = Vo * sin

= 25 * sin ()

= 25 * 0.656059

= 16.4014 m/s

a) At the top of its trajectory, its vertical component is momentarily zero, leaving only the horizontal component that you just figured. So that will be the velocity at the top of its trajectory.

Va = Vox = 18.86773 m/s

b) its given half that height

so find that half

H = Voy^2 / 2g

= (16.4014)^2 / 2*9.81

=13.7108 m

Half that height is H/2 = 6.8554 m

so we have the formula for distance as

here initial velocity is u = Voy and a= g ( as g will be acting downwards)

When you throw a ball in the air the velocity will be same at each point of when it is ascending and descending.

so lets take distance travelled s = H/2 we are talking in terms of vertical only

as we have already established that horizontal component wont be bothering the flight

so substituting these and finding the time required to go H/2

H/2 = Voy * t + (1/2 *(-g) * t^2)

6.8554 = 16.4014 * t - 0.5 * 9.81 *(t^2)

4.905 (t^2) -16.4041 * t + 6.8554 = 0

by solving this quadratic equation we find t = 2.854 s (only this root satisfies the equation)

now find the vertical velocity at t = 2.854 s by using , here a = - g as it is in the opposite direction of the velocity

Vy = Voy - gt

= 16.4014 - (9.81 * 2.854)

= -111.596 m/s

this is in the upward direction while descending it will be in the downward direction so it will be

+11.596 m/s

so now total velocity will be

Vb =

Vx = Vox = 18.86773 m/s and Vy = 11.596 m/s

Vb = = 22.146 m/s

c)

Now calculate the percentage

(Vb/Va) * 100 = (22.146/18.86773) * 100 = 117.37%

The Stone is moving 17.37 % faster in part b than in part a