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kellyzukes · Oct 4, 2009, 01:14 PM

This seems so easy to me but I just can't figure it out..

A ball is thrown vertically upward with a speed of 26.4 m/s.
How high does it rise?
Answer in units of m.

Perito · Oct 4, 2009, 02:19 PM

$D = V_0T + \frac 12 AT^2$

D = distance traveled
Vo = initial velocity (26.4 m/s)
T = time (in seconds because of the units of Vo
A = acceleration, in this case, the acceleration due to gravity in m/sec

Make sure you use the right sign on the acceleration (usually negative for gravity).

ebaines · Oct 5, 2009, 06:14 AM

The formula Perito suggsted doesn't quite do it, as you would need to know what value for T (time) to plug into his equation. You can find that from:

$<br /> V = V_0 + AT<br />$

Set V = 0 (because that's the velocity of the ball at the top of its trajectory) and then solve for T. This gives the time to reach the top of tracetory. Remember that the value for the ball's acceleration is -g. Once you determine the value f T, you can put it into Perito's equation and solve for D.

However, an easier way to solve this is to use the equation:

$<br /> V^2 - V_0^2 = 2AD <br />$

Set V to 0 and solve for D.

Both of these methods should give you the same answer. I suggest you try it both ways to confirm.