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

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:

Where:

is the final height = 0

is the initial height = 0

is the initial vertical velocity of the stone

t: is the time

g: is the gravity = 9.81 m/s²

By solving the above quadratic equation we have:  

t = 6.62 s

b) The maximum range is:

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

I hope it helps you!

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

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