On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 18 m/s at an angle 45 degrees above the horizontal.
a) How much farther did the ball travel on the moon than it would have on earth? (answer in m)
b) For how much more time was the ball in flight?
To start off, list what you know, it'll help you find a way to find what you're looking for. You know that
y-initial = 0m ax= 0 xf=?
x-initial = 0m ay=1/6[-9.81m/s(squared)] yf=0m (it lands back on the surface)
Vix=? t=? Vfx=?
Viy=? Vfy=?
Vi=18m/s
First, split the velocity into its x-component and y-component
Vix=cos45(Vi)
Viy=sin45(Vi)
now, you have everything you need to use yf=yi +Viyt + 1/2ayt(squared) to find time (on the moon)
after you've found the value of time, use the same equation for the x-component to find xf, which will be where the ball lands relative to the initial point. (on the moon)
xf = xi + Vixt +1/2axt(squared)
Now, on Earth, the initial velocities (both components), as well as x-initial, and y-initial and y-final will be the same as they were on the Moon. However, ay will now be
-9.8m/s(squared). So use that in the yf=yi +Viyt + 1/2ayt(squared) equation to find time (on Earth) and compare it to the time on the Moon. Then, using that time in that equation's x-component counterpart, find xf (on Earth) and compare that to the xf you calculated on the Moon.
Hope this helps, good luck!