Question

A cannon tilted up at a 35.0 ∘ angle fires a cannon ball at 82.0 m/s from atop a 13.0 m -high fortress wall. What is the ball's impact speed on the ground below? Express your answer with the appropriate units.

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the cannon ball = V₁ = 82 m/s

Angle the cannon ball is fired at = = 35^o

Initial horizontal velocity of the cannon ball = V_1x = V₁Cos = (82)Cos(35) = 67.17 m/s

Initial vertical velocity of the cannon ball = V_1y = V₁Sin = (82)Sin(35) = 47.03 m/s

Height of the fortress wall = H = 13 m

Time taken by the cannon ball to reach the ground = T

When the cannon ball reaches the ground the vertical displacement of the cannon ball is directed downwards therefore it is negative.

T = 9.857 sec or -0.268 sec

Time cannot be negative.

T = 9.857 sec

Velocity of the cannon ball's when it reaches the ground = V₂

Horizontal velocity of the cannon ball when it reaches the ground = V_2x

There is no horizontal force acting on the cannon ball therefore the horizontal velocity of the cannon ball remains constant.

V_2x = V_1x

V_2x = 67.17 m/s

Vertical velocity of the cannon ball when it reaches the ground = V_2y

V_2y = V_1y + gT

V_2y = 47.03 + (-9.81)(9.857)

V_2y = -49.67 m/s

V₂ = 83.54 m/s

Ball's impact speed on the ground = 83.54 m/s