Projectile motion- acceleration due to gravity on the moon

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SUMMARY

The discussion focuses on calculating the distance a golf ball hit by astronaut Alan Shepard on the Apollo 14 mission traveled on the moon compared to Earth, given the moon's gravity is 1/6 that of Earth's. The ball was struck at a speed of 18 m/s at a 45-degree angle. Key calculations involve using projectile motion equations to determine the time of flight and horizontal distance traveled on both the moon and Earth. The suggested approach includes decomposing the initial velocity into its x and y components and applying the kinematic equations for both environments.

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Homework Statement


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?


Homework Equations





The Attempt at a Solution



I don't really know where to start at all, any hints for what to do would be greatly appreciated!

a)
18cos45= 12.73
18sin45= 12.73
 
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Do you know how to figure out how long it would be in flight if it was hit straight upward at a given speed?
 
Nope, I've never taken physics before, I'm pretty lost.

Would this be it?
y = y(i) + v(i)*sin(theta)*t + 1/2*g*t^2
 
Last edited:
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!
 

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