# Finding the initial speed of two balls.

1. Jul 20, 2014

### zim70094

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

Ball A is thrown straight upward and returns to its original level in 2 seconds. Ball B is thrown at a 40 degree angle above the horizontal and also returns to its original level in 2 seconds. What is the initial speed of each ball.

2. Relevant equations

ΔV=at
ΔX=Vit+1/2at^2
v^2-Vi^2=2a(ΔX)
ΔX=1/2(V+Vi)t

3. The attempt at a solution
for ball A I replaced acceleration with gravity and tried to find Vi with the first equation a=-9.81 and t=2s but I didn't know what to do with Vf. I thought maybe it was zero and came up with 19.62 m/s but i know that's not right.

for ball b i know that Vi in the y direction is Visinθ and x direction is Vicosθ but i'm not positive how to apply this information. I think I may have to use my knowledge of vectors too but i'm not sure. for example i think I use V=square root of (Vix^2 + Viy^2)

2. Jul 20, 2014

### voko

When a ball is thrown straight upward, there is a indeed a certain point where its speed is zero. But where is that?

3. Jul 20, 2014

### zim70094

the max height

4. Jul 20, 2014

### Nathanael

What must the two situations (straight up and 40° angle) have in common?

5. Jul 20, 2014

### zim70094

the max height? i'm not sure.

6. Jul 20, 2014

### zim70094

Would I have to cut t (2s) in half?

7. Jul 20, 2014

### Nathanael

What determines how long an object is in the air?

8. Jul 20, 2014

### zim70094

i'm not sure. gravity?

9. Jul 20, 2014

### Nathanael

Good, what else?

If I throw a ball perfectly horizontally it doesn't stay in the air very long, does it?

Why is that?

10. Jul 20, 2014

### zim70094

because gravity is acting in the direction toward the ground...

11. Jul 20, 2014

### Nathanael

Ok but what makes it last longer when I throw it upwards?

How would you calculate the time the object is in the air?

12. Jul 20, 2014

### zim70094

Having an initial velocity. Please stop trying to raise your post count and provide a reasonable solution. I'm just trying to refresh myself on the subject and this was a problem I wasn't positive about.

13. Jul 20, 2014

### zim70094

I'm down for helping me get to the solution, but you are being ridiculous.

14. Jul 20, 2014

### Nathanael

You're not positive about the problem, and that's WHY I'm trying to help you understand. (As opposed to giving you the full solution in a single post....)

They BOTH have an initial velocity. What is different about the initial velocities? The VERTICAL component...

The time in the air depends on gravity and the vertical component of the velocity... Therefore both the "straight up throw" and "40° throw" must have the same vertical component of velocity.

P.S.
I could care less about my post count, but thanks for trying to be rude......

15. Jul 20, 2014

### Nathanael

People understand things better when they discover it themself.

I could tell you right off the bat that they both need to have the same vertical component, but what's the point in telling you that if you don't understand why....

16. Jul 20, 2014

### zim70094

so I can use the two equations to find the missing variables? what does that have to do with max height?

17. Jul 20, 2014

### Nathanael

It doesn't directly have to do with max height, (I probably shouldn't have quoted you on my first reply) but two objects have the same max height if and only if they have the same initial vertical velocity (assuming they're launched from the same height).

And two objects have the same time in the air if and only if they have the same max height (assuming same launch height and gravity and what not)

I don't know what two equations you're talking about but I'm sure you have it right

18. Jul 21, 2014