- #1
blaxican707
- 5
- 0
Homework Statement
A ball is thrown into the air with an initial vertical velocity of 5.0m/s in addition to its horizontal velocity of 25.0m/s, how far would the ball travel horizontally before it reached it's peak?
Homework Equations
vf = vi + at
I've also tried vf = 1/2g +at
The Attempt at a Solution
Ok, this is where I made numerous attempts, but this one seems to be more logical.
vf = vi + at
0 = 5.0m/s + (-9.8m/s^2)t
I put the vi for the inital velocity for vertical, then I inputed -9.8 because of the velocity being countered by the gravitational pull.
then I rearange to -9.8m/s^2 by itself by doing
0 + (9.8m/s^2)t = 5.0m/s
then i divide both sides by 9.8m/s^2, looking like this
[tex]\frac{9.8m/s^2}{9.8m/s^2}[/tex] = [tex]\frac{5.0m/s}{9.8m/s^2}[/tex]
which would then leave me with t = .51 seconds
with that i would multiply that by the horizontal distance, looking like
.51 seconds x 25m/s = 12.8m
since 12m is the distance, i divide that by 2, giving me 6.4m reaching its peak.
If this is too hard to read, my bad, I'm not use to inputting equations on a board. Thanks for the help in advance. And was this format correct according to the sticky posted in this forum?
Last edited: