Simple Projectile Motion

  • Thread starter LiiArch
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  • #1
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Homework Statement


A projectile is fired at 57.5 deg above the horizontal. Its initial speed is equal to 175.0 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile? At what time being after being fired does the projectile reach this maximum height?


Homework Equations


Y(t) = Y0 + Vy0t - 1/2at^2


The Attempt at a Solution


I already figured out that the maximum height would be 1110 m. So I thought I could plug that into the equation above and figure out t.

1110 = 0 + 142.5t - 4.905t^2

How do I solve this? :/
 

Answers and Replies

  • #2
21
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You have an equation already; are you asking how to solve for t?

It's a quadratic equation...

EDIT: That equation doesn't have any real solutions; are you sure that 142.5 is the initial velocity in the y-direction?
 
  • #3
10
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Yes. That's why I'm so confused.

I enter 1.11 x 10^3 for the first part and it reads back correct, so 1110 is the max height.

Let me double check the y velocity.

sin(57.5) = Vy/V0
sin(57.5) = Vy/175.0
175.0*sin(57.5) = Vy
147.6 = Vy

Why do I keep doing this? I need to double check better.

EDIT: 1110 = 147.6t - 4.905t^2

Solutions for this come out to 15.32615142 and 14.7655917. These both come up as incorrect.
 
Last edited:
  • #4
gneill
Mentor
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EDIT: 1110 = 147.6t - 4.905t^2

Solutions for this come out to 15.32615142 and 14.7655917. These both come up as incorrect.

Keep some additional decimal places in your intermediate results, and round at the end. Quadratics can sometimes be 'sensitive' that way. Round the result to an appropriate number of significant figures; some software is picky about this!

To make your life a bit easier, you might also want to recognize that the maximum height achieved represents the halfway point of the trajectory; the projectile will take just as much time to fall from that height as it did to reach that height. So the result can also be found by considering the time it would take for an object dropped from that height to hit the ground. If h is the distance from the release point to the ground, then [itex] h = \frac{1}{2}g t^2 [/itex]. Solve for t.
 
  • #5
I dont get my answer yet plz somebody helps me...
 

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