Projectile motion problem

  • #1
OP warned about not using the homework template
A ball is shot from the ground into the air. At a height of 9.1 m, its velocity vector (in unit vector notation) is (7.6i - 4.9j) m/s; that is, vx = 7.6 m/s and vy = -4.9 m/s.

What was the maximum height of of the ball above the ground?

I swear to god i have been working on this for 2 hours and got nowhere! >=[. i tried
-4.9 = -9.8(t)+vy0 and solved for t. i then plugged t into the position formula and got some crazy answer that makes no sense! Please give me a better understanding of this problem!!
 

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  • #2
haruspex
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4.9 = -9.8(t)+vy0 and solved for t.
How could you solve for t without knowing vy0?
Or do you mean you expressed t as a function of vy0 and plugged that into the vertical height formula?
Please post your working.
 
  • #3
BvU
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Hello Isu, :welcome:

Do not erase the template. It's there for you too. Sort out your equations and variables ! Knowns, unknowns, etc.

You can't solve for t if you don't know vy,0
 
  • #4
BvU
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Haha ... help is on the way. A reprimand from a mentor too !
 
  • #5
hey guys i solved it but I'm stuck on another part. and yes i expressed vyo as an equation with t in it. i tried a different approach and got it. now I'm stuck again.
 
  • #6
haruspex
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hey guys i solved it but I'm stuck on another part. and yes i expressed vyo as an equation with t in it. i tried a different approach and got it. now I'm stuck again.
So please post what that part asks for and show your working as far as you get.
 
  • #7
A ball is shot from the ground into the air. At a height of 9.1 m, its velocity vector (in unit vector notation) is (7.6i - 4.9j) m/s; that is, vx = 7.6 m/s and vy = -4.9 m/s.

What was the maximum height of of the ball above the ground?

i answered 10.3 meters. it was correct. i solved this by setting the time = 0 to that 9.1 meter height and solving. made it much simpler
------------------------------------------------------------------------------------------------------------------------------------------
What is the speed of the projectile at its maximum height?

7.6m/s, i solved this knowing that the vy component was 0 there and the vx component was still constant
------------------------------------------------------------------------------------------------------------------------------------------------
At the moment the ball is 9.1m above the ground, how far away is the ball from its launch point on the ground?

i answered 14.8 meters. correct
------------------------------------------------------------------------------------------------------------------------------------------------
At the moment the ball is 9.1m above the ground, how much farther will the ball travel before it returns to the ground?
now here i am stuck. i started a new integral starting at the apex and changing the sign of the acceleration. i started the height at 10.3 meters. so my position equation is -4.9t^2+10.3 and the starting vy was 0 because it started at the maximum. now i am getting the answer 14.8 and it says its wrong
 
  • #8
i plugged in 1.4 seconds to the vx position and got that answer. so it says FROM that 9.1m position. the x position was 14.8 there. so i took the total displacement from there which was 14.8
 
  • #9
Hello Isu, :welcome:

Do not erase the template. It's there for you too. Sort out your equations and variables ! Knowns, unknowns, etc.

You can't solve for t if you don't know vy,0
thank you
 
  • #10
haruspex
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started a new integral
Integral?
starting at the apex
Why not just carry on from the 9.1m height?
changing the sign of the acceleration
The acceleration is the same throughout.
now i am getting the answer 14.8 and it says its wrong
Hard to tell what you did without seeing all your working.

Edit... But I notice you got that it was the same as the distance back to the launch point. Do you see what that suggests regarding your error? It looks like you have effectively run time backwards, or took the vertical velocity at 9.1m height as up instead of down.
 
  • #11
Integral?

Why not just carry on from the 9.1m height?

The acceleration is the same throughout.

Hard to tell what you did without seeing all your working.

Edit... But I notice you got that it was the same as the distance back to the launch point. Do you see what that suggests regarding your error? It looks like you have effectively run time backwards, or took the vertical velocity at 9.1m height as up instead of down.
I see.. I will make another attempt when I get home. Thank you. At work right now
 
  • #12

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