Posted by Haley on .
King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 42 m/s and at an angle of 33°. A cannonball that was accidentally dropped hits the moat below in 1.6 s.
(a) How far from the castle wall does the cannonball hit the ground?
(b) What is the ball's maximum height above the ground?
Use the dropped cannonball's fall time (1.6 s) to compute the height of the top of the castle.
H = (g/2)t^2 = 12.54 m
(a) In this case, assume they are talking about the fired cannonball, not the dropped one.
Solve this equation for time of flight t:
H = 12.54 -(g/2)t^2 + 42sin33 t = 0
X = 42cos33*t
b) Ymax = H + (42sin33)^2/(2g)
Note that the castle height H had to be added to the distance that the cannonball actually rises.