Question

Question #295c7

Answer 1

The cannonball will land 236.25m far from the ship.

Explanation:

Since we ignore any friction for this problem, the only force applying to the cannonball is its own weight (it's a free fall). Therefore, its acceleration is:

`a_z = (d^2z)/dt^2 = -g = -9.81 m*s^(-2)`

`rarr v_z(t) = dz/dt = int((d^2z)/dt^2)dt = int(-9.81)dt`
`= -9.81t+v_z(t=0)`

Since the cannonball is fired horizontally, `v_z(t=0)=0 m*s^(-1)`

`rarr v_z(t) = -9.81t`

`z(t) = int(dz/dt)dt = int(-9.81t)dt = -9.81/2t^2+z(t=0)`

Since the cannonball is fired from a height of 17.5m above the sea level, then `z(t=0) = 17.5`

`z(t) = -9.81/2t^2+17.5`

We want to know how long it will take the cannonball to reach the ground:

`z(t) = -9.81/2t^2+17.5 = 0`

`rarr t = sqrt(17.5*2/9.81) = sqrt(35/9.81) ~~ 1.89s`

Now, we want to know how far the ball can go during this time. Since the ball was fired at an initial speed of `125m*s^(-1)` with no resistance, then:

`d = v*t = 125*1.89 = 236.25m`