Question

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where does the projectile hit the ground and with what speed?

Answer 1

The projectile's horizontal and vertical positions at time
`t` are given by

`x=\left(250(\rm m)/(\rm s)\right)\cos45^\circ\,t`

`y=30\,\mathrm m+\left(250(\rm m)/(\rm s)\right)\sin45^\circ\,t-\frac g2t^2`

where
`g=9.8(\rm m)/(\mathrm s^2)`. Solve
`y=0` for the time
`t` it takes for the projectile to reach the ground:

`30+(250)/(\sqrt2)t-4.9t^2=0\implies t\approx36.2458\,\mathrm s`

In this time, the projectile will have traveled horizontally a distance of

`x=(250(\rm m)/(\rm s))/(\sqrt2)(36.2458\,\mathrm s)\approx6400\,\mathrm m`

The projectile's horizontal and vertical velocities are given by

`v_x=\left(250(\rm m)/(\rm s)\right)\cos45^\circ`

`v_y=\left(250(\rm m)/(\rm s)\right)\sin45^\circ-gt`

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about
`√(176.77^2+(-178.43)^2)(\rm m)/(\rm s)\approx250(\rm m)/(\rm s)`.