Question

King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 60 m/s and at an angle of 27°. A cannonball that was accidentally dropped hits the moat below in 1.0 s. (a) How far from the castle wall does the cannonball hit the ground? Incorrect: Your answer is incorrect. m (b) What is the ball's maximum height above the ground? m

Answer #1

s = 1/2 a t² = 4.9 * 1.0² = 4.9 meters.

The horizontal velocity is constant and equal to 60 * cos (27°) =
53.5 m/s.

The initial vertical velocity is 60* sin (27°) = 27.2 m/s. This is
the same speed (in reverse) that the ball would attain if dropped
after t = v / a = 27.2/ 9.8 = 2.78 seconds.

The maximum height above the wall is therefore 1/2 a t² = 4.9 *
2.78² = 37.87 meters, and above the moat 37.87 + 4.9 = 42.77
meters.

This height of 42.77 meters = 1/2 a t², so t² = 42.77 * 2 / a =
85.54/9.8 = 8.73, and t = 2.95 seconds. Add 2.78 seconds for a
total time of 5.73 seconds, multiply this by the horizontal
velocity of 53.5 m/s.

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