# Question #295c7

##### 1 Answer
Jun 12, 2015

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:

${a}_{z} = \frac{{d}^{2} z}{\mathrm{dt}} ^ 2 = - g = - 9.81 m \cdot {s}^{- 2}$

$\rightarrow {v}_{z} \left(t\right) = \frac{\mathrm{dz}}{\mathrm{dt}} = \int \left(\frac{{d}^{2} z}{\mathrm{dt}} ^ 2\right) \mathrm{dt} = \int \left(- 9.81\right) \mathrm{dt}$
$= - 9.81 t + {v}_{z} \left(t = 0\right)$

Since the cannonball is fired horizontally, ${v}_{z} \left(t = 0\right) = 0 m \cdot {s}^{- 1}$

$\rightarrow {v}_{z} \left(t\right) = - 9.81 t$

$z \left(t\right) = \int \left(\frac{\mathrm{dz}}{\mathrm{dt}}\right) \mathrm{dt} = \int \left(- 9.81 t\right) \mathrm{dt} = - \frac{9.81}{2} {t}^{2} + z \left(t = 0\right)$

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

$z \left(t\right) = - \frac{9.81}{2} {t}^{2} + 17.5$

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

$z \left(t\right) = - \frac{9.81}{2} {t}^{2} + 17.5 = 0$

$\rightarrow t = \sqrt{17.5 \cdot \frac{2}{9.81}} = \sqrt{\frac{35}{9.81}} \approx 1.89 s$

Now, we want to know how far the ball can go during this time. Since the ball was fired at an initial speed of $125 m \cdot {s}^{- 1}$ with no resistance, then:

$d = v \cdot t = 125 \cdot 1.89 = 236.25 m$