Question

king arthurs knights fire a cannon from the top of the castle wall. the cannonball is fired at a speed of 50m/s and an angle of 30 degree. A cannonball that was accidentally dropped hits the moat below in 1.5s.

How far from the castle wall does the fired cannon hit the ground?

Answer 1

Answer:

Explanation:

initial height = 0.5*gt^2 = 0.5*9.8*1.5 = 7.35 m

now,

x - x0 = ut + 1/2at^2

0 - 7.35 = 50sin(30)*t + (1/2)*(-9.8)*t^2

4.9t^2 - 25t - 7.35 = 0

solving this and taking only the positive value of t

t = (sqrt(76906)+250)/98 or 5.3808 secs

a)

s = 50*cos(30 degree)*(sqrt(76906)+250)/98

s = 232.9959 m or 234 approx.