Question

An object is thrown up in the air at 15 meters per second. You want to figure out how high the object will be at 2.5 seconds after its release.

Use the 5 motion variables and list the values and directions of the variables already known. Use a motion equation to find the height of the object.

My answer:
t is 2.5 s
Vi is 0 m/s
Vf is 15 m/s up
a is -9.8 m/s^2 up
d is to be found
Which equation should I use?

Answer 1

t is 2.5
Vi is 15 m/s up
a = -9.8 m/s^2 up
(or, 9.8 m/s^2 down)

d = Vi * t + a/2 t^2
= 15*2.5 - 4.9 * 2.5^2 = 6.875 m

Vf is Vi + at = 15 - 9.8 t^2
(though it does not matter for this question)

Answer 2

Thanks!

So the final velocity is not a single, known value?

Also how do I find out if the object is still going up or coming down?

Answer 3

it stops going up when v=0. So, since

v = 15 - 9.8t
v=0 at t=1.53

So, at t=2.5 it is coming back down:

http://www.wolframalpha.com/input/?i=15t+-+4.9t%5E2