Question

A huge shell is fired from an old navy warship towards a pirate ship. If the shell is fired horizontally at a height of 25.0 meters from the water with a velocity of 75.0 m/sec
A) How long will the shell take to hit the water?

B) At what distance should the firing ship close to sink a shell into the pirate ship at a height of 5.00 meters above the pirate ship's water line?

Answer 1

The shell's vertical position $y$ at time $t$ is given by

$y=25.0\,\mathrm m-\dfrac g2t^2$

where $g=9.81\,\dfrac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. The shell hits the water when $y=0$, so the time it takes for that to happen is

$0=25.0\,\mathrm m-\dfrac g2t^2\implies t=2.26\,\mathrm s$

For the shell to reach a height of $y=5.00\,\mathrm m$, it must travel for a time $t$ such that

$5.00\,\mathrm m=25.0\,\mathrm m-\dfrac g2t^2\implies t=2.02\,\mathrm s$

The shell's horizontal position $x$ is given by

$x=\left(75.0\,\dfrac{\mathrm m}{\mathrm s}\right)t$

so that after 2.02 seconds, the warship should be at a distance

$x=\left(75.0\,\dfrac{\mathrm m}{\mathrm s}\right)(2.02\,\mathrm s)=153\,\mathrm m$

away from the pirate ship in order to hit it at the desired height.