Question

A cannon ball is fired from the top of a 56 meter cliff at a velocity of 22 meters per second.

A) How far from the base of the cliff does the cannon ball land?

B) What is the actual velocity of the cannon ball at impact?

C) What is the angle of impact?

Answer 1

We assume that the cannon ball is fired horizontally.

So horizontal velocity of the ball = vx = u = 22 m/s

vertical component of velocity = vy = 0 + g t 
   or,   (vy)² = 0² + 2 g h 
                   = 2 * 10 m/s² *56 m
                   = 1120 m²/s²

Velocity just before impact with the ground = v
   v² = 1120 + 22²   (m/s)²
   v = 40.04 m/s

Time taken to fall from the top of the cliff to the ground = t sec.
     h = u t + 1/2 g t² 
     56 = 0 + 1/2 * 10 * t²
     t² = 56/5 
     t = 3.346 sec.

Distance traveled horizontally = vx * t = 22 * 3.346 = 73.61 m

Vertical component of velocity just before hitting  the ground 
    = vy = √1120  m/s = 33.466 m/s
 
Angle of impact = angle made by the velocity with the ground
    = Tan⁻¹ (33.466 m/s) / (22 m/s)
    =  56.68°

Answer 2

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