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Confusing projectile motion problem - need guidance

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Confusing projectile motion problem -- need guidance

Here is the picture of the diagram.
333441200.jpg


The Question:
A woman throws a ball at a vertical wall d = 3.6 m away. The ball is h = 1.8 m above ground when it leaves the woman's hand with an initial velocity of 16 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)
(a) Where does the ball hit the ground?
m (away from the wall)

(b) How long was the ball in the air before it hit the wall?
s

(c) Where did the ball hit the wall?
m (above the ground)

(d) How long was the ball in the air after it left the wall?
s


What I have done so far
Ok, this problem is very complicated for me so I am going step by step, so if anyone can guide me through this, Im grateful.

The triangle made from the person throwing the ball, the hyp. is 16, I figured out the adjacent and the opposite by use sin45*16 and cos45*16 which is 11.3.

For Question A How long was the ball in the air before it hit the wall?
I am having trouble conceptualizing this, will the velocity be 0 before it hits the wall? if it is, I can do change in velocity = A*T and that ='s to 1.6s but how do I find if this is correct? Or there is something I am not seeing.
 

Answers and Replies

  • #2
Doc Al
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The triangle made from the person throwing the ball, the hyp. is 16, I figured out the adjacent and the opposite by use sin45*16 and cos45*16 which is 11.3.
OK. Those are the components of the initial velocity.
For Question A How long was the ball in the air before it hit the wall?
I am having trouble conceptualizing this, will the velocity be 0 before it hits the wall?
No.

Hint: Analyze the horizontal motion.
 
  • #3
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I dont know if this is correct but to calculate horizontal range of a projectile is multiplying the x component * time.

I have 3.6m = 11.3m/s * t where t = 3.1s
 
  • #4
Doc Al
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I have 3.6m = 11.3m/s * t where t = 3.1s
Your equation is correct, but you solved for t incorrectly.
 
  • #5
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Oh lol, is the other 3.6/11.3.

For the next question
(a) Where does the ball hit the ground?

It mentioned that the horizontal component of velocity is reversed.
So it is -11.3 m/s?

Will the velocity slow down because it hit the wall or because it mentioned it is reversed it still stays constant 11.3m/s

Any hints on how to start this?
BTW thanks for helping me.
 
  • #6
LowlyPion
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Oh lol, is the other 3.6/11.3.

For the next question
(a) Where does the ball hit the ground?

It mentioned that the horizontal component of velocity is reversed.
So it is -11.3 m/s?

Will the velocity slow down because it hit the wall or because it mentioned it is reversed it still stays constant 11.3m/s

Any hints on how to start this?
BTW thanks for helping me.
For this part you might want to ignore the wall since it only reverses and doesn't reduce the speed. First solve for total distance with no wall and then sub out the difference caused by the wall.
 
  • #7
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This is what I did.

I wanted to find out the time and I am calculating
change in velocity = A * T where change in velocity = 0m/s-11.3m/s = -9.81m/s * T
which = T = 1.2s and 2T = 2.4s.

Then I used horizontal range equation
X = 11.3m/s(2.4s) = 27.12m
then subtract 3.6m = 23.52m

Is this approach correct?
 
  • #8
LowlyPion
Homework Helper
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This is what I did.

I wanted to find out the time and I am calculating
change in velocity = A * T where change in velocity = 0m/s-11.3m/s = -9.81m/s * T
which = T = 1.2s and 2T = 2.4s.

Then I used horizontal range equation
X = 11.3m/s(2.4s) = 27.12m
then subtract 3.6m = 23.52m

Is this approach correct?
The first part is right but ...

You need to find the Total time it is in the air. (Since the wall doesn't affect the speed except to reverse direction you can do that in 2 parts.)

The first part of determining time is to figure time to Maximum height. And that you did correctly. But you can't double it.

The return trip will need to travel back to the same level plus the height h it was released from to get to the ground level. It needs to be accounted for.

The time to earth from maximum is 1/2 a t2 = x + h = x + 1.8 m
 
  • #9
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If the ball from initial position to the maximum height in 1.2s,
The reason I double the time is because from the maximum height falling back to the initial position will take another 1.2s. Or in this case is not valid?


1/2(-9.81m/s)(1.2s)^2 = x + 1.8m
(-4.905m/s)(1.44s) = x + 1.8m
-7.06m = x + 1.8m
x = -8.86m

Is this correct for a negative distance?
 
  • #10
LowlyPion
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If the ball from initial position to the maximum height in 1.2s,
The reason I double the time is because from the maximum height falling back to the initial position will take another 1.2s. Or in this case is not valid?


1/2(-9.81m/s)(1.2s)^2 = x + 1.8m
(-4.905m/s)(1.44s) = x + 1.8m
-7.06m = x + 1.8m
x = -8.86m

Is this correct for a negative distance?
The way the equation was set up was Total X is equal the 7.06 and the 1.8.

Now knowing the Total distance you can figure the Time from the same equation -
Total X = 1/2 at2
 
Last edited:
  • #11
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X = 8.86m

8.86m = 1/2(9.81m/s)t^2

8.86m = 4.905m/s * t^2
1.8s = t^2
t = 1.3s

X = 11.3m/s*1.3s
X = 14.7m

14.7m - 3.6m = 11.1m
Is that the answer?
 
Last edited:
  • #12
LowlyPion
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X = 8.86m

8.86m = 1/2(9.81m/s)t^2

8.86m = 4.905m/s * t^2
1.8s = t^2
t = 1.3s

X = 11.3m/s*1.3s
X = 14.7m

Is that the answer?
Not quite.

The total time then is the time to go up and the time to drop.

That would be 2.5 s right? (1.2 + 1.3)

That puts Total horizontal distance at (11.3) (2.5).

But that distance includes the distance to the wall and the return from the wall. And that is 7.2 m. When you subtract that now you have the distance BEHIND the woman throwing the ball that it lands.

EDIT: They are asking for distance from the wall so just subtract the 3.6 m.
 
Last edited:
  • #13
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I thought we are calculating the time at return from the wall to the ground.
Because I already calculated the time from the person to the wall which was .32s.

The question asked
(a) Where does the ball hit the ground?
m (away from the wall)

away from the wall so we shouldn't need to subtract right?
11.3*2.5 would give us 28.25m away from the wall since we are calculating the time from the return from the wall?
 
  • #14
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Not quite.

The total time then is the time to go up and the time to drop.

That would be 2.5 s right? (1.2 + 1.3)

That puts Total horizontal distance at (11.3) (2.5).

But that distance includes the distance to the wall and the return from the wall. And that is 7.2 m. When you subtract that now you have the distance BEHIND the woman throwing the ball that it lands.

EDIT: They are asking for distance from the wall so just subtract the 3.6 m.
Oh, thanks so much.


For
(c) Where did the ball hit the wall?
m (above the ground)

(d) How long was the ball in the air after it left the wall?
s


Is D 2.5seconds? since we used that for Question A?

For C - I have to find the distance and + 1.8m since it is thrown 1.8m above ground.

Since vertical velocity is 11.3m/s and the time we got before was .32seconds
so 11.3m/s * .32s = 3.61 m

Do I add 1.8m to that since the question ask m above ground?
 
  • #15
LowlyPion
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Oh, thanks so much.

For
(c) Where did the ball hit the wall?
m (above the ground)

(d) How long was the ball in the air after it left the wall?

Is D 2.5seconds? since we used that for Question A?
No. Read the question more carefully.

For C - I have to find the distance and + 1.8m since it is thrown 1.8m above ground.

Since vertical velocity is 11.3m/s and the time we got before was .32seconds
so 11.3m/s * .32s = 3.61 m

Do I add 1.8m to that since the question ask m above ground?
The time you determined to get to the wall is how you will determine it ... indirectly.

First use that time (.32 s) to determine what the vertical speed was when you get to the wall.
V = Vo - a t = 11.3m/s - 9.8 (.32s)

Then by knowing the difference in velocity, you can use
Vf2 - Vi2 = 2 a x.
And that is your x height that you add to 1.8 m.
 
  • #16
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Wow thanks so much. You helped me get this far!

Is it 2.5sec - .32sec?
 
  • #17
LowlyPion
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Wow thanks so much. You helped me get this far!

Is it 2.5sec - .32sec?
For part d, yes that would be correct.
 
  • #18
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Thanks Alot LowlyPion!!!
 
  • #19
LowlyPion
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Thanks Alot LowlyPion!!!
No problem. What I hope you have figured out is how to keep the vertical and horizontal components separate, and how to use what you know about the events in one direction that help you determine things in the other.

Good luck.
 

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