Question

During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +48.5" m"/s at an angle of 42.0°, how long does it take the stone to hit the ground? For those settings, what is the maximum range? How high will the stones go? Show all your work

Answer 1

Answer:

a) t = 6.62 s

b) x = 238.6 m

c) H = 53.7 m

Explanation:

a) We can find the time of flight as follows:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ is the final height = 0

$y_{0}$ is the initial height = 0

$v_{0_{y}}$ is the initial vertical velocity of the stone

t: is the time

g: is the gravity = 9.81 m/s²

$v_{0}sin(42)t - \frac{1}{2}gt^{2} = 0$      

$48.5 m/s*sin(42)*t - \frac{1}{2}9.81 m/s^{2}*t^{2} = 0$  

By solving the above quadratic equation we have:  

t = 6.62 s

b) The maximum range is:

$x = v_{0_{x}}t = 48.5 m/s*cos(42)*6.62 s = 238.6 m$

c) The maximum height (H) can be found knowing that at this height the final vertical velocity of the stone is zero:

$v_{f_{y}}^{2} = v_{0_{y}}^{2} - 2gH$  

$H = \frac{v_{0_{y}}^{2} - v_{f_{y}}^{2}}{2g} = \frac{(48.5 m/s*sin(42))^{2} - 0}{2*9.81 m/s^{2}} = 53.7 m$

I hope it helps you!