# Projectile motion, finding initial speed

Tags:
1. Mar 15, 2015

### vizakenjack

An object fired at an angle of 30° above the horizontal takes 2.5 s to travel the last 12 m of its vertical distance and the last 10 m of its horizontal distance. With what speed was the object launched? (Note: This problem does not stipulate that the initial and final elevation of the object has to be the same!)

Solution:
First, an initial velocity in x direction (air friction is ignored, I guess, so velocity in x direction stays the same) is to be found.

The object traveled in x direction - 10 meters.
And the formula for horizontal distance is: d = vt
10 = v*2.5s
v = 4m/s
that's the x component of velocity (Vx)

Knowing initial Vx and an angle, we can find the y component of the velocity (Vy)
Tangent is opposite/adjacent. We need to find opposite, which corresponds to Vy.
so: tan(30°) * Vx (adjacent) = Vy
Vy = 2.309 m/s

Now the initial velocity needs to be found. Using Pythagoras theorem: sqrt(4^2 m/s + 2.309^2 m/s) =
4.618602 m/s

But my question is, why doesn't the solution work if instead of finding initial horizontal velocity, initial vertical velocity is to be found first?
See for yourself:
So I see 12 meters is the max height reached by an object, right?
Formula for height in projectile motion is: h = V2y/2g -> 12=V2y/2g
Vy = 15.3362 m/s
Why??

I'm also kinda confused whether I should use this formula:
d = V*t + 1/2 * gt2
Anyway, using that kinematic equation yields: 17.05 m/s for Vy
Which is also wrong...

Last edited: Mar 15, 2015
2. Mar 15, 2015

### Misha Kuznetsov

I'm not sure, but it seems to me that you are finding the initial velocity in the x direction for a situation where the total horizontal distance is 10. Is that what the problem is asking? It seems to me that the horizontal distance is unknown and you only know the time for the last 10 meters.

3. Mar 15, 2015

### vizakenjack

The problem gives distance in both y and x direction.
However, the problem is solvable if you first find the Vx.

Why can't I solve it finding Vy first and then solving for Vx and Velocity?

4. Mar 15, 2015

### Cake

The first equation can't be used because it uses the initial velocity to or maximum height to solve it. You have neither those things (hence the note in the problem). The second equation requires knowledge of the initial position in the y direction. Which you also don't have. The values you're given don't qualify.

5. Mar 15, 2015

### vizakenjack

Isn't the "12 m of its vertical distance" = max height reached by an object?

Wouldn't the initial y position be 0? Or why can't I assume it's 0?

6. Mar 15, 2015

### Cake

1. No. It doesn't say in the problem that is the max height.
2. No, it tells you this in the note in the problem.