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-\frac g2t^2`

where
`g=9.81\,(\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-\frac 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-\frac g2t^2\implies t=2.02\,\mathrm s`

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

`x=\left(75.0\,(\mathrm m)/(\mathrm s)\right)t`

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

`x=\left(75.0\,(\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.