How to find distance and velocity with just acceleration and time

[SignSeeker7]

Supposing an object is thrown into the air and the only relative data you know is the amount of time it was in the air, the mass of the object, and the acceleration due to gravity being -9.8 m/s^2, can you figure out the distance it traveled and the velocity it had when it landed? And how? (We're also ignoring air resistance and assuming that the object was thrown from and caught at the same height.)

 

[the_emi_guy]

Yes you can.
Look up "kinematic formulas" on the Internet. Hopefully you can take it from there, maybe report back what you discover.

 

[Janus]

Yes, assuming it is thrown straight up. Otherwise, you would have to know what angle it was thrown at.

 

[SignSeeker7]

I'm having trouble working with the equations I've found because I don't know the initial and final velocities (after being thrown upward and before being caught) or the distance.
I have:

V_{}f = V_{}i + a*t

V_{}f² = V_{}i² + 2*a*d

d = V_{}i*t + \frac12{}{}*a*t²

d = ((V_{}i + V_{}f) / 2) * t

What exactly would I do? There's probably a pretty obvious solution, but I just don't see it.

 

[Ken G]

I'll assume the object is thrown straight up. The trick is to notice a symmetry in the motion-- the path upward to a stop, and the path downward from that stop, are just the time-reversed versions of each other. This means if you know the total time, how would you figure out just the time to go up? Then what do you know about the velocity at the top, given that this is at the top of the motion? Calling the velocity at the top V_f, how can you use the first equation you wrote above to relate the time you figured out to the V_i you want to know? And once you know V_i, how will that tell you the final velocity when the object comes back down? The symmetry is the key.

In fact, if you don't want to go through these steps (informative though they are), your first equation can be used all in one fell swoop, if you simply recognize the relationship that the symmetry implies about V_i and V_f, where now you can interpret them as the initial and final velocities of the entire motion (hint: signs matter, apropos to your handle).

 

[SignSeeker7]

Wow. What's really sad is that I was thinking about how they were symmetrical, but I never bothered to actually split it in half and solve it that way. Thanks a lot, Ken G!