# Homework Help: Projectile motion

1. Sep 18, 2009

### tigerlili

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

You throw a ball toward a wall at speed 24.0 m/s and at angle θ0 = 41.0° above the horizontal (Fig. 4-35). The wall is distance d = 21.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall?

2. Relevant equations

v0x= v0costheta
v0y=v0sintheta

y(t) = -1/2at^2 + v0y + y0

3. The attempt at a solution

this one seems like it should be easy but i'm having a lot of trouble
i got the correct v0x, it was 24cos41= 18.11
but the computer is saying that 24sin41= 15.75 is incorrect.. so i can't get the v0y component

also.. i just don't know how to do part a
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 18, 2009

### Staff: Mentor

They are looking for the velocity components when it hits the wall, not v0y. The y-component is accelerated.

Hint: You'll need expressions for x(t) & y(t).

3. Sep 18, 2009

### tigerlili

so.. i'm still confused
do you think you could explain to me how to get the accelerated version of the y component?

i know these are probably stupid questions.. but i feel like i haven't really learned this properly :/

4. Sep 18, 2009

### Staff: Mentor

The y-component has constant acceleration. What kinematic formula relates velocity to time for accelerated motion?

5. Sep 18, 2009

### tigerlili

d= vit + 1/2 at^2 ?

6. Sep 18, 2009

### Staff: Mentor

No, that gives position as a function of time; you need velocity as a function of time. (But you'll need that equation for part a, which you need to solve first to get the time.)

7. Sep 18, 2009

### tigerlili

okay, so..

for part a i use the equation i just mentioned, using v0x as vi and 21=d to solve for time?

and then i plug in the time into the equation... vf=vi + a*t solving for vf?

8. Sep 18, 2009

### Staff: Mentor

Yes. What's the acceleration in the horizontal direction? (That equation simplifies.)

Yes, for the vertical component.

9. Sep 18, 2009

### tigerlili

i wish this were working for me..but it's not :(

Last edited: Sep 18, 2009
10. Sep 18, 2009

Hi there tigerlili! Don't give up yet!

If you post all of your steps and calculations along the way
it will be a lot easier for us to see where you are
making mistakes

11. Sep 18, 2009

### Staff: Mentor

Show what you tried and we'll set you straight. (Don't give up now. )

12. Sep 18, 2009

### tigerlili

i did say what i was doing.. it's just a couple of posts up
(sorry to sound so distraught.. but my physics prof literally teaches us nothing :/)

i tried to use d= vit + 1/2 at^2 for part a but then i realized i was solving for time, not distance, and i got really confused, even though i was told that i was doing it correctly:

for some reason it won't let me quote what i wrote, so, here it is:

Re: projectile motion

Originally Posted by tigerlili
"for part a i use the equation i just mentioned, using v0x as vi and 21=d to solve for time?"
Yes. What's the acceleration in the horizontal direction? (That equation simplifies.)

"and then i plug in the time into the equation... vf=vi + a*t solving for vf?"
Yes, for the vertical component.

13. Sep 18, 2009

### Staff: Mentor

Those steps are correct, but you must have made an error somewhere if you're not getting the answer.

Did you solve for the time? What's the acceleration in the horizontal direction? (That's a trick question. Or an easy question, depending on how stressed out you are!)

And you'll use that time to solve for Vy. (You'll also use that time to solve for the height where the ball hits the wall.)