Can somebody help me double check this?

You are on the surface of the moon and throw a rock straight up. Height after t seconds will be: h(t) = V0*t -1.625t^2 meters.

V0 = initial velocity.

Q: When does the rock hit the ground/ surface of the moon? (answer is a value of t dependent on V0)

Thanks

Question

Can somebody help me double check this?

You are on the surface of the moon and throw a rock straight up. Height after t seconds will be: h(t) = V0*t -1.625t^2  meters.

V0 = initial velocity.

Q: When does the rock hit the ground/ surface of the moon?  (answer is a value of t dependent on V0)


Thanks

mertsj · Answer

$$V _{o}-1.625t^2=0$$
$$-1.625t^2=-V _{_{o}}$$
$$t^2=\frac{V _{o}}{1.625}$$
$$t=\sqrt{\frac{V _{o}}{1.625}}$$

anonymous · Answer

huzzah. We match. Thanks.

mertsj · Answer

yw

anonymous · Answer

Actually, this kept bothering me. Should be: t = v0/1.625  ...

h(t) = V0*t -1.625t^2  ...

1. set equation equal to zeo 
2. Factor out a t ...t(V0 - 1.625t) ... t = v0/1.625

mertsj · Answer

Oh gees.  I didn't see that t.  Yes.  Just like you have it.