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Projectile motion

  • Thread starter mimsteel
  • Start date
  • #1
7
0

Homework Statement


A baseball was thrown at an angle of 45degrees above the horizonal, it traveled a horizontal distance of 296ft. and it was caught at the same level from which it was thrown. Neglect air resistance. A.) What was the ball's initial speed? B.) how long was the ball in the air.


The Attempt at a Solution


Ok heres where I got to.

I used V^2=Vi^2+2a(90.2m) (296ft to 90.2m)

This gives me a Vi of 42.05 m/s correct??

Now I take that 42.05m/s and get the X component and get 29.73 m/s and use formula

X=ViT+.5T^2

This gives me
90.2=29.73 T+ .5T^2

So, Ill have T^3= (29.73+90.2)*2 ??

Giving me a time of 6.21 Seconds.


SO T=6.21s

and Vi=42.05m/s
even though i use 29.73 m/s in the time equation 42.05 is my initial velocity correct??
 
Last edited:

Answers and Replies

  • #2
7
0
If someone could tell me what formula to use so i can start to try to figure this out. Not looking for a full solution just where to start off
 
  • #3
123
1
Well, the 45 degrees is a big hint.
 
  • #4
7
0
Ya like i split it into X and Y components. And i have that. I just dont have a time. So im not sure which formula to use from there. So like X=209.3 and the Y= 209.3 as well.
 
  • #5
123
1
It would be helpful if you list all the known and unknowns first.
Here:
[tex]x=269[/tex]
[tex]y=0[/tex]
[tex]v_{x}=?[/tex]
[tex]v_{1}=?[/tex]
[tex]t=?
[/tex]

y equals to 0, because it came back to where it started- vertical displacement is 0.
 
Last edited:
  • #6
7
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So, I need to solve for time before i can get the Vi correct?
 
  • #7
123
1
Not necessarily. It would be helpful if you know that with three unknowns, you just need three equations, and you are done.
 
  • #8
7
0
Ok heres where I got to.

I used V^2=Vi^2+2a(90.2m) (296ft to 90.2m)

This gives me a Vi of 42.05 m/s correct??

Now I take that 42.05m/s and get the X component and get 29.73 m/s and use formula

X=ViT+.5T^2

This gives me
90.2=29.73 T+ .5T^2

So, Ill have T^3= (29.73+90.2)*2 ??

Giving me a time of 6.21 Seconds.


SO T=6.21s

and Vi=42.05m/s
even though i use 29.73 m/s in the time equation 42.05 is my initial velocity correct
 
Last edited:
  • #9
7
0
Can anyone verify i worked that properly??
 
  • #10
123
1
There is no acceleration in the x direction. Therefore you use uniform motion formulas. i.e.
[tex]x=v_{x}t[/tex]

And now let me show this to you separately.
For horizontal direction:
[tex]x=269[/tex]
[tex]v_{x}=?[/tex]
[tex]t=?[/tex]
Notice that there aren't v1 or v2, because they are unchange, there's no acceleration.

For vertical direction:
[tex]y=0[/tex]
[tex]v_{1}=?[/tex]
[tex]v_{2}=-v_{1}[/tex] Make sure you know why this is.
[tex]t=?[/tex]
[tex]a=-g[/tex]
For vertical motion, a is -g which is constant, therefore, you use uniform acceleration formulas.
 

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