Projectile Motion help

  • Thread starter AlexT89
  • Start date
  • #1
3
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Homework Statement



This is a projectile motion problem, providing only the maximum height the object reaches and the angle that it leaves the ground, and asking to find the speed at which the object must leave in order to reach that max height:

The object leaves the ground at 45 degrees and reaches a max height of 12 feet. With what speed must the object leave the ground to reach that height?

Homework Equations



y-direction:
where Voy = Vo sin 45,
y displacement (12 feet) = Voyt + (0.5)(-9.8m/s2)t2

Vo = (Voy) / (sin 45)

The Attempt at a Solution



I feel like I'm going in circles every time I attempt to solve this problem, almost like I don't have enough information, even though I know that I do.

I've gotten as far as this:

3.66 m = (Vo)(sin 45)t - 4.9t2

I will appreciate any help you can give me. I think I am looking at it wrong, but I can't figure out how else to do so.

Thanks in advance,
Alex
 

Answers and Replies

  • #2
69
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When projectile reaches it's highest point, shouldn't the vy be equal to zero?
 
  • #3
3
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When projectile reaches it's highest point, shouldn't the vy be equal to zero?

Right, so if I used the equation:

Vy2 = Voy2 + 2(-9.8)(y-disp)

it would allow me to solve for Voy, and by extension Vo.
Is that correct?
 
  • #4
3
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That worked out, I got the answer that it gives in the back of the book. I knew I was overlooking something. Thanks for your help, method man.
 
  • #5
69
0
Right, so if I used the equation:

Vy2 = Voy2 + 2(-9.8)(y-disp)

it would allow me to solve for Voy, and by extension Vo.
Is that correct?
Yes something like that. Yours y-disp is maximum height and of course Vy=0.

You could also do it like this:
vy=v0y-g*t
And when height is maximum, vy=0 so you get v0y=g*t. From here, you can extract either t or v0y and put it in that first equation: y = Voyt + (0.5)(-9.8m/s2)t2.
 

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