# Projectile motion (ball thrown horizontally)

1. Dec 30, 2008

### boomer77

1. The problem statement, all variables and given/known data
a ball is thrown horizontally from a height of 16.01m and hits the groung with a speed that is 5.0 times its initial speed. what was the initial speed?

2. Relevant equations

3. The attempt at a solution

dx=?
vx=?
t=?

dy= -16.01m (because the ball is falling that distance)
viy= vi
vfy= 5vi
a= -9.8 m/s^2

(5vi)^2= vi^2+ 2(-9.8)(-16.01)

25 (vi^2)= vi^2+ 313.8

what should i do next if this is right?

2. Dec 30, 2008

In the y direction, (Vy)i=0 because it was thrown horizontally. So it only had an initial horizontal component of velocity.

When it lands, it has both a horizontal and vertical component of velocity.

Now, with that in mind, can you see an equation that will let you find the final y-component of velocity?

Hint: you have dy, ay and (Vy)i.

3. Dec 31, 2008

### boomer77

i am following you so far

so my work will now look like

(5vi)^2=0^2+2(-9.8)(-16.01)
25(vi^2)=313.796
divided by 25
vi^2=12.55
vi=3.54

i plugged this into my problem set and it says it is wrong, so do wil i take this answer and put it into another equation? or is my math wrong?

4. Dec 31, 2008

### Staff: Mentor

Solve for vi, of course! Try first solving for vi^2, then taking the square root.

5. Dec 31, 2008

### boomer77

i did do that and my answer came up as 3.55 (rounded) but my problem set is still saying it's wrong. where is my error?

vf^2=vi^2+2(-9.8)(-16.01)
(5vi)^2=0+313.8
25(vi^2)=313.8
313.8/25= 12.6
vi^2=12.6
vi=3.55

6. Dec 31, 2008

### boomer77

this equation that you wrote

17.8^2+3.6^2=(5x3.6)^2

what is it? i'm pretty new to physics (i started taking physics a this trimester at my highschool) could you explain how to solve it?

7. Dec 31, 2008

### Staff: Mentor

Your solution in your first post was correct. That's the one I want you to finish.

8. Dec 31, 2008

### boomer77

but isn't that 3.55 my answer?

9. Dec 31, 2008

### Staff: Mentor

No. You changed your equation, making it incorrect. Go back to your first post, or just see my first comment.

10. Dec 31, 2008

### Kindayr

You're so close :)

For questions like these you have to break everything into components, solve them, then put them all back together.

What I think you're doing wrong is breaking them into components, then using a mixture of components and other data within the wrong equation.

Here's a hint: Pythorgean Theorem

(I apologize for simply solving it before, hope this helps

11. Dec 31, 2008

### boomer77

so is this 3.55 one component?

12. Dec 31, 2008

### Staff: Mentor

As I've pointed out (several times), your initial solution was perfectly correct. (You might not realize that it was correct.) Why don't you finish it?

What allows you to combine perpendicular components is the fact that you are really combining their squares: V² = (Vx)² + (Vy)²

13. Dec 31, 2008

### Kindayr

It would be if its magnitude was correct.

Think of it this way:

At the beginning, the ball's actual velocity has two components; an x valued velocity, and a y valued velocity.

At the end, the ball's actual velocity is composed of two parts just like in the beginning.

Think Pythagorean Theorem.

14. Dec 31, 2008

### boomer77

ohh i get it

my problem will be 3.55^2+x^2=5x^2!
then solve for x!