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${}^{{\mathbf{v}}_{\mathbf{0}}\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$ and at an angle of${}^{{\mathbf{u}}_{\mathbf{0}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}}$. 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)?

It is given that,

$$\begin{gathered} v_0=28\mathrm{m}/\mathrm{s} \\ \mathrm{\theta }=40° \end{gathered}$$

To find the speed at the top of the wall, find vertical and horizontal components of velocity at top of the parabolic path, from these two components, find resultant speed at top. As there is no acceleration along horizontal direction, so horizontal speed remains constant. Acceleration due to gravity is always downwards, so vertical speed decreases and becomes zero at maximum height.

Formulae:

The displacement is given by,

$y=y_0+v_{0}t+\frac{1}{2}\times a{t}^2$ (i)

The final velocity is given by,

role="math" localid="1654243677233" $v_f=v_0+at$ (ii)

$v_f^2=v_0^2+2a∆y$ (iii)

Where${}^{∆y}$ is the resultant displacement

Consider the velocity in horizontal and vertical direction as ${}^{v_x}$and${}^{v_y}$respectively. As there is no acceleration along horizontal direction, there would be no change in horizontal speed, i.e. horizontal speed remains constant.

So, at the top, horizontal speed is

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=28\times \cos \left(40\right) \\ =21.45\mathrm{m}/\mathrm{s} \end{gathered}$$

And at the top, the y component of the velocity${}^{v_y}$is zero as acceleration due to gravity is in downward direction,Thus, the speed of the stone is given by,

$$\begin{gathered} v=\sqrt{v_x^2+v_y^2} \\ v=\sqrt{21.{45}^2+0^2} \\ v=21.45m/s \end{gathered}$$

Vertical component of initial velocity is given as,

${\mathrm{v}}_{0\mathrm{y}}=28\times \sin 40=18\mathrm{m}/\mathrm{s}$

Now, at maximum height,

$$\begin{gathered} v_f^2=v_{oy}^2+2\times a\times y_{top} \\ 0={18}^2-2\times 9.81\times y_{top} \end{gathered}$$

So,

${\mathrm{y}}_{\mathrm{top}}=16.51\mathrm{m}$

Now, half of maximum height is${}^{\mathrm{y}=8.26\mathrm{m}}$

Now, vertical speed at half height is,

role="math" localid="1654244962738" $$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{0\mathrm{y}}^2+2\mathrm{ay} \\ {\mathrm{v}}_{\mathrm{f}}^2={18}^2-2\times 9.81\times 8.26 \\ {\mathrm{v}}_{\mathrm{f}}=12.7\mathrm{m}/s \end{gathered}$$

And, the horizontal component of velocity at half of height isrole="math" localid="1654245031748" ${}^{21.45\mathrm{m}/s.}$

So, speed at half of height is,

$$\begin{gathered} \mathrm{v}=\sqrt{21.{45}^2+12.7^2} \\ \mathrm{v}=24.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, percentage difference is,

$v\%=\frac{24.9-21.45}{21.45}=16.3\%$

Therefore, at the top, vertical velocity is zero. Horizontal velocity remains constant because there is no acceleration along horizontal direction.