Question

King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 40 m/s and at an angle of 35°. A cannonball that was accidentally dropped hits the moat below in 1.2 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?

Answer 1

a)

t = time of travel = 1.2 s

consider the motion in horizontal direction

V_ox = initial velocity inhorizontal direction = 40 Cos35 = 32.8 m/s

a_x = acceleration = 0 m/s²

X = distance from the castle wall

Using the equation

X = V_ox t + (0.5) a_x t²

X = (32.8) (1.2) + (0.5) (0) (1.2)²

X = 39.4 m

b)

consider the motion of ball dropped

Y_o = initial position = h = height of wall

Y = final position = 0

a_y = acceleration = - 9.8 m/s²

t = time = 1.2 s

V_oy = initial velocity = 0 m/s

Using the equation

Y = Y_o + V_oy t + (0.5) a t²

0 = h + 0 (1.2) + (0.5) (- 9.8) (1.2)²

h = 7.056 m

consider the motion of ball launched along Y-direction

V_fy = final velocity at maximum height = 0

V_oy = initial velocity = 40 Sin35 = 22.94 m/s

a = acceleration = - 9.8

Y_o = initial position = 7.056 m

Y_max = maximum height

Using the equation

V_fy² = V_oy² + 2 a (Y_max - Y_o)

0² = 22.94² + 2 (-9.8) (Y_max - 7.056)

Y_max = 33.91 m