Height of a projectile at a certain point

  • Thread starter ridik88
  • Start date
  • #1
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Homework Statement


A soccer player kicks a soccer ball at a speed of 33.5 m/s at an angle of 10.6 degrees at the goal 25.0m away.


Homework Equations



What is the height of the ball when it reaches the goal?

The Attempt at a Solution



Was able to calculate the final velocity but need a little help setting up this one
 

Answers and Replies

  • #2
radou
Homework Helper
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Come on, please show your work.
 
  • #3
1,341
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[tex]d = 1/2(v_o+v_f)t[/tex]
[tex]d = v_{o}t+1/2at^2[/tex]
[tex]v_f = v_o+at[/tex]
[tex]v_{f}^2 = v_{o}^2+2ad[/tex]

Those equations are a good place to start^^

Remember there's no acceleration in the "horizontal" direction
 
Last edited:
  • #4
5
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Work so far

Ok... so far I calculated

sr = (2(33.5 m/s)^2)/(9.81 m/s^2) cos10.6 sin10.6

sr = 41.4m

this will give me the range if the ball struck the goal at the same height it was kicked from. I cant figure out how to calculate its height 25.0m away.
 
  • #5
1,341
3
I don't see how finding the total range of the ball will help you...

[tex]d = v_{o}t+1/2at^2[/tex]

If you tried to solve this for the horizontal direction then [tex]a=0 m/s^2[/tex]. So you're left with [tex]d = v_{o}cos(10.6)t[/tex]. You have a displacement (the goal) and you know the intial velocity so you can figure out how long it will take to reach the goal...
 
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  • #6
5
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I calculated the time to be 1.71s, when plugged into the equation to solve for "d" it gives me a height of 10.35m. Looks like a feasible answer (hope I did it right lol). Thanks so much for helping.
 
  • #7
hage567
Homework Helper
1,509
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10.35 m?!? That's pretty high! How did you arrive at that value for t?
 
  • #8
5
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Vf = Vo + at

22.99m/s = 6.16m/s + 9.81m/s^2 ( t )

t = 1.71s
 
  • #9
1,341
3
You don't know what the velocity will be when the ball reaches the goal unless you previously know the time. Where did you get 22.99 m/s from?
 
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  • #10
5
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[tex]v_{f}^2 = v_{o}^2+2ad[/tex]

Vf^2 = (6.16m/s)^2 + 2(9.81m/s^2)(25m)

Vf = 22.99m/s

wrong way I guess
 
  • #11
87
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The key to projectile questions are knowing which formula to use, and 9 times out of ten, you need to work out the horizontal components before you can start on the veritical (due to the effect of acceleration on vertical velocity over time).

As we have estbalished however, the is a link between the two .. time.

List out what you know about the hoiztonal components, and choose a formula that will be easiest to find time with.

Then you can start working out the vertical elements.
 
  • #12
orb
6
0
The key is that the horizontal component of motion is at constant velocity, and that the vertical component is at constant acceleration (due to gravity). So separate your two components. Using [tex] x = v_{x} t [/tex] solve for t and then use that in the equation for uniform acceleration.
 

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