Question

An object is dropped 14ft above the surface of the moon. How long will it take the object to hit the surface of the moon if d^2/dt^2=-5.2 ft/sec^2? Or see attachment.

Answer 1

d^2s/dt^2 is simply the second derivative of a position function that defines acceleration on the moon. Do this problem the same way you would with a projectile on Earth, except instead of g = 9.8m/s^2, it will be -5.2ft/sec^2

Answer 2

May you walk me through this problem? :X

Answer 3

I don't have the required kinematic equations memorized. I'm sure they're somewhere in your book, rearranged from v = v(initial) + at, or x-x(initial) = etc etc. Just use -5.2ft/sec^2 as your acceleration and 14ft as your distance, d

Answer 4

Umm I'm guessing it's this one. :X
xf = xi + vi*t + 1/2*a*t^2

Answer 5

yes, exactly, where your x(final) is the final position, -14ft, and x(initial) is 0
a = -5.2ft/sec^2, solve for time (initial velocity is 0 so you can just get rid of that term and not worry about quadratics)

Answer 6

0 = 14ft + 0*t + 1/2*(-5.2ft/s^2)*t^2

Answer 7

Theeeen... solve for t?

Answer 8

Yep, that's it. looks like... -28 = -5.2ft/s^2*t^2, so the sqrt(-28/5.2)

Answer 9

sorry the negatives will cancel out so you wont have a negative in the square root sqrt

Answer 10

Aaaaah okay. So the final answer it sqrt(-28/5.2)? :D

Answer 11

Yes, without the negative. If you have trouble solving systems of equations you should review, it won't go away \[\sqrt{28/5.2}\]

Answer 12

Haha, sure will. It's a practice problem that I saw in the book. Got a bit confused, that's all. Thanks though. :)