Question

A cannonball at ground level is aimed 30 degrees above the horizontal and is fired with an initial speed of 125 m/s. A) how far away from the cannon will the cannonball hit the ground? B) What is the highest point of its trajectory?

Answer 1

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

Gravitational acceleration = g = 9.81 m/s²

Initial velocity of he cannonball = V₀ = 125 m/s

Angle the cannonball is fired at = = 30^o

Initial horizontal velocity of the cannonball = V_x0 = V₀Cos = (125)Cos(30) = 108.25 m/s

Initial vertical velocity of the cannonball = V_y0 = V₀Sin = (125)Sin(30) = 62.5 m/s

Time taken by the cannonball to come to the ground again = T

When the cannonball reaches the ground again the vertical displacement of the cannonball is zero.

0 = V_y0T + gT²/2

0 = V_y0 + gT/2

0 = 62.5 + (-9.81)T/2

T = 12.74 sec

Horizontal distance covered by the cannonball = R

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

R = V_x0T

R = (108.25)(12.74)

R = 1379 m

Maximum height reached by the cannonball = H

Vertical velocity of the cannonball at maximum height = V_y1 = 0 m/s

V_y1² = V_y0² + 2gH

(0)² = (62.5)² + 2(-9.81)H

H = 199.1 m

A) Distance the cannonball hits the ground from the cannon = 1379 m

B) Highest point of the trajectory of the cannonball = 199.1 m