Question

A cannonball is launched horizontally off a 80 m high castle wall with a speed of 25 m/s. How long will the cannonball be in flight before striking the ground

Answer 1

2.23 s.

Step-by-step explanation:

Given:

S = 80 m

voy = 0 m/s

vox = 25 m/s

g = 9.81 m/s^2

Using equation of motion,

S = vo*t + 1/2 * g * t^2

80 = 0 + 4.905t^2

Solving this equation, we have:

t = sqrt(16.31)

= 4.039 s

Therefore, t = 4.04 s.

Answer 2

The cannonball will be spend 4.04 seconds before striking the ground.

Step-by-step explanation:

Given:

vertical height = 80 m

horizontal velocity = 25m/s

Initial vertical velocity, Vy= 0

Applying equations of motion;

s = Vy*t + ¹/₂gt²

80 = ¹/₂*9.8t²

4.9t² = 80

t² = 80/4.9

t² = 16.327

`t = √(16.327) =4.04 s`

Therefore, the cannonball will be spend 4.04 seconds before striking the ground.