Projectile Motion problem. very lost.

  • Thread starter Chip90
  • Start date
  • #1
55
0
[SOLVED] Projectile Motion problem. very lost.

Homework Statement



A person is traveling in an accelerating car. He throws a ball in the direction the car is acceleration with a velocity of vo. In the final condition, he catches the ball. Note the car begins accelerating after the ball is thrown.

What is the time the ball is in the air in terms of vox and a?

2. Equations

Projectile motion ones?

The Attempt at a Solution



Well, the ball and the person have the same a. Also, x (the distance traveled) and t (time) are the same.

I have no idea how to setup this up or what approach to use.

F=ma doesn't really help, it just comes down to a=-g.

TIA

I attached a pic. in case my description doesn't help
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    9.5 KB · Views: 343
Last edited:

Answers and Replies

  • #2
136
0

Well, the ball and the person have the same a. Also, x (the distance traveled) and t (time) are the same.


x and t are, indeed the same. However, the ball and the person do not have the same acceleration. What happens to the ball's acceleration as soon as the ball leaves the person's hand?
 
  • #3
6
0
are we assuming that the ball is thrown perfectly horisontal with no variables such as air resistance?
Try these equations:
v=u+at
v(squared)=u(squared)+2as
S=ut+0.5at(squared)

where S is displacement, a is acceleration,v is final velocity, and u is initial velocity.
 
  • #4
55
0
the balls acceleration is gravity, which is in y direction?

so person has a in x direction
ball has -g in y direction

so how do i link these? what equations. i really appreciate your help.
 
  • #5
6
0


F=ma doesn't really help, it just comes down to a=-g.



g has no affect on the horisontal componant but if the ball was thrown upwards into an arc, then it would have to be considered using simple trig
 
  • #6
55
0
p=person b=ball

Xp = 0.5AT^2

Xb=0.5AT^2 + VoxT

Yb=-0.5GT^2 + VoyT=0

Also, schoolboy, there is no air resistance. The ball is thrown with an angle from horizontal. According, to p21bass, we cant use your equations or mine since ab doesn't equal ap. is this right?

fyi the ball is thrown at angle between horizontal and vertical.
 
  • #7
6
0
can you tell me the numbers to this equation?
 
  • #9
6
0
what is the angle?
 
  • #10
55
0
Angle is not given. There are no numbers. You have to solve for time in terms of Vox (for the ball) and a(acceleration of the person).
 
  • #11
6
0
I'll get back to you on this
 
  • #12
55
0
ok thanks. i need to figure this out by 7 lol
 
  • #13
136
0
Ignore the y-component. And we don't need the angle. They are both completely unnecessary for this problem. What is the x-component of acceleration of the ball, after it leaves the person's hand? Once you realize what it is, what equations can you use?


EDIT: It's also important to know if the car/person is accelerating at a constant acceleration.
 
Last edited:
  • #14
55
0
ok the Vox of the ball is Vo * cos thetha?

The car is accelerating at constant acceleration.

I know the set of equations for projectile motion, but I am not sure what I need to solve this problem.

in post 6, are at least the first 2 i posted correct for this problem?
 
  • #15
2
0
Hey

The ball you throw will not have an acceleration in the x direction since the only force acting on it is gravity, which affects it only on the y axis, so the formula for the displacement of x is Δx = (Vox)(Δt)

The car starts accelerating but at the beginning it is at rest, so its formula for x displacement is Δx = 0 + (1/2)(ax)(Δt)

For him to be able to catch the ball from his car, their Δxs need to be equal so his hand and the ball are at the same place, so make both equations equal each other:

(Vox)(Δt) = (1/2)(ax)(Δt^2)

That's it.

From looking at the problem I assume its not thrown at an angle, otherwise it would say in terms of theta as well. However, if it did, the equation would be:

cos(θ)(Vo)(Δt) = (1/2)(ax)(Δt^2)

Understand why the cos(θ) is there? Just to get the x velocity even if it was thrown at an angle. This is of course assuming it doesn't hit the ground before the driver can catch it, but I think that assumption is built into the problem.

Hope it helps
 
Last edited:
  • #16
136
0
Again - we don't care about the angle, or the y-component or gravity. We are only concerned with what's going on in the x direction. Since you know that the distance traveled is the same for both the ball and person, you can set two equations (one each for distance traveled) equal and solve for t that way. Also, since you know that the person doesn't start accelerating until he's thrown the ball, you know what v_0 is. And think about what the acceleration is in the x-direction for the ball. You'll see that a couple of terms simply drop out, and you're left with two equivalent equations for distance - one of the ball, and one of the person.
 
  • #17
55
0
That does help, but once you boil down to that equation, wouldn't the time just cancel? i need to find it in terms of T.
 
  • #18
136
0
From looking at the problem I assume its not thrown at an angle, otherwise it would say in terms of theta as well. However, if it did, the equation would be:

cos(θ)(Vox)(Δt) = (1/2)(ax)(Δt)

This is wrong. v_ox = v_o*cos(theta).
 
  • #19
55
0
This is wrong. v_ox = v_o*cos(theta).

just caught that. also the time would cancel in this eq right?
 
  • #20
136
0
No. One of Shaun's equations is wrong, as well - the right-hand one, specifically.
 
  • #21
55
0
Again - we don't care about the angle, or the y-component or gravity. We are only concerned with what's going on in the x direction. Since you know that the distance traveled is the same for both the ball and person, you can set two equations (one each for distance traveled) equal and solve for t that way. Also, since you know that the person doesn't start accelerating until he's thrown the ball, you know what v_0 is. And think about what the acceleration is in the x-direction for the ball. You'll see that a couple of terms simply drop out, and you're left with two equivalent equations for distance - one of the ball, and one of the person.

well i think it is this

for ball

x=Vox*T

for person

x= 0.5*A*T^2.

is that correct?

so

Vox*T=0.5*A*T^2

2*Vox=A*T

and thus

T= 2*Vox/A

is that right?
 
  • #22
136
0
YES!!! Now, since x is the same for both, just set the two equations equal to each other and solve for t.
 
  • #23
136
0
Yeah, that's it. It's one of those problems that seems tougher than it really is.
 
  • #24
2
0
Whoops, I forgot to multiply the acceleration by (1/2)(t^2). I typed it in wrong.

sorry about that

the velocity of the ball formula was right, though. The problem does not mention any angle so its assumed its x velocity.
 
  • #25
55
0
Thanks guys, that was confirmed to be correct!
 

Related Threads on Projectile Motion problem. very lost.

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
4
Views
740
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
536
  • Last Post
Replies
11
Views
871
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
987
  • Last Post
Replies
13
Views
2K
Top