Question

6. A cannon with a barrel velocity of 140 m/s launches a cannonball horizontally from a tower. Neglecting air resistance, how far vertically will the cannonball have fallen after 4 seconds?

Answer 1

Answer:

The approximate distance traveled by the ball would be 80 meters in 4 seconds.

Explanation:

Given the canon has a horizontal velocity 140 m/s when it is launched from a tower.

We can see that as was fired horizontally the vertical component of the velocity is zero.

Now, the only force that causes ball to fall to the ground is the gravitational force.

Also, it was given that we need to neglect air resistance.

So, this the condition of free fall. We will use its formula

$S=\frac{1}{2} gt^2$

Where,

$S$ is the distance traveled vertically.

$t$ is the time of fall.

$g$ is  $9.81\ m/s^2$.

Plugging these values we get,

$S=\frac{1}{2}\times 9.81\times 4^2=78.48\ m$

So, the approximate distance traveled by the ball would be 80 meters in 4 seconds.