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

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

$y=v_0\sin\theta\,t-\dfrac g2t^2$

where $v_0=45\,\dfrac{\rm m}{\rm s}$, $\theta=30^\circ$, and $g=9.80\,\dfrac{\rm m}{\mathrm s^2}$. The ball reaches the ground when $y=0$ at

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

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

At this time, the ball's horizontal position is

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

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

$\dfrac{\left(45\,\frac{\rm m}{\rm s}\right)^2\sin60^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\approx180\,\rm m$