# Projectile horizontal motion Question

1. Jan 29, 2008

### blaxican707

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

vf = vi + at

I've also tried vf = 1/2g +at

3. 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 its self by doing

0 + (9.8m/s^2)t = 5.0m/s

then i divide both sides by 9.8m/s^2, looking like this

$$\frac{9.8m/s^2}{9.8m/s^2}$$ = $$\frac{5.0m/s}{9.8m/s^2}$$

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: Jan 29, 2008
2. Jan 29, 2008

### hage567

You have two vertical velocities?!? Which is supposed to be the horizontal one? I need to know that first.

3. Jan 29, 2008

### blaxican707

my bad, the 5.0m/s is the vertical velocity and the 25.0m/s is the horizontal velocity!

4. Jan 29, 2008

### hage567

You mean horizontal velocity, right?
You don't need to divide by two here. When you found t by saying the vertical velocity (vf) is zero (this is at the peak height), you essentially did this, since this is the halfway point of the ball's trajectory. Does that make sense?

This equation is wrong. Where did you get it?

5. Jan 29, 2008

### blaxican707

Yes horizontal velocity. So the ball reached its horizontal peak at 12.8m? I thought it was 6.4 because I had originally thought that 12.8m was the total distance the ball had traveled untill it hit the ground. -9.8m/s^2 in the equation had already gave me the vertical peak correct?

I looked at my notebook and wrote down the wrong formula :(

6. Jan 29, 2008

### hage567

If you think of the trajectory of the ball, the vertical velocity will be zero (momentarily) at the peak height.
Since you used vf = vi - at, and set vf =0 and solved for t what you found was the time for the ball to get to the peak height. Since the trajectory is symmetric here, this is equal to half the horizontal distance.

Another way you could have done it is to have used $$y - y_0 = v_0 t + (1/2) a t^2$$

where y - y0 is the vertical displacement.

Here, set the vertical displacement equal to zero, since the ball returns to the ground. Solve this one for t. What do you notice? If you put this value for t into your equation for the horizontal distance, what do you get? This is where you would need to divide by two.

See http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra4

Hope that helps.

7. Jan 30, 2008

### blaxican707

I'm trying to solve that equation right now... kinda of confusing.

But back to the original post - ok so if the ball reached it's peak at a horizontal distance of 12.8m, then if the trajectory of the ball is symetrical, the ball traveled 25.6m when it hit the ground - correct?

8. Jan 30, 2008

Yes this looks correct. If the ball reached its vertical peak (neglecting air resistance this coincides with its midpoint) at a horizontal distance of 12.8m, then the total horizontal distance is 25.6.

9. Jan 30, 2008

### blaxican707

ok great, I was looking at this as common sense - but then agian this is phsyics and did'nt think that this answer would be that simple.

I appreciate the help