Question

A golf ball is struck at 30 degrees above the horizontal with a velocity of 45m/s. How far horizontally will the ball travel? The answer is 180m, but I don’t know how to get that answer.

Answer 1

The ball's horizontal position
`x` and vertical position
`y` at time
`t` are given by

`x=v_0\cos\theta\,t`

`y=v_0\sin\theta\,t-\frac g2t^2`

where
`v_0=45\,(\rm m)/(\rm s)`,
`\theta=30^\circ`, and
`g=9.80\,(\rm m)/(\mathrm s^2)`. The ball reaches the ground when
`y=0` at

`0=v_0\sin\theta\,t-\frac g2t^2=\left(v_0\sin\theta-\frac g2t\right)t=0\implies t=\frac{2v_0\sin\theta}g`

(we don't care about
`t=0`)

At this time, the ball's horizontal position is

`v_0\cos\theta\left(\frac{2v_0\sin\theta}g\right)=\frac{{v_0}^2\sin2\theta}g`

which you might recognize as the range formula. With the known parameters, the ball thus traverses a range of

`(\left(45\,(\rm m)/(\rm s)\right)^2\sin60^\circ)/(9.8\,(\rm m)/(\mathrm s^2))\approx180\,\rm m`