Projectile motion (ball thrown horizontally)

  • Thread starter boomer77
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  • #1
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Homework Statement


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?


Homework Equations





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

vf^2= vi^2 + 2ad

(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?
 

Answers and Replies

  • #2
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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
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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
Doc Al
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vf^2= vi^2 + 2ad

(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?
Solve for vi, of course! Try first solving for vi^2, then taking the square root.
 
  • #5
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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
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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
Doc Al
Mentor
45,011
1,288
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
Your solution in your first post was correct. That's the one I want you to finish.
 
  • #8
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but isn't that 3.55 my answer?
 
  • #9
Doc Al
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but isn't that 3.55 my answer?
No. You changed your equation, making it incorrect. Go back to your first post, or just see my first comment.
 
  • #10
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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
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
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so is this 3.55 one component?
 
  • #12
Doc Al
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45,011
1,288
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
161
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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
33
0
ohh i get it

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

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