Find Initial Velocity Given Height, Range, and Initial Angle

  • Thread starter vcm1992
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  • #1
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Homework Statement


So! The problem states that a grasshopper jumps from the edge of a table. We know that x-initial is zero and x-final is 1.06m. This is how far he jumped from the origin. The height of the table is unknown, but we know that the height of the grasshoppers jump with respect to the table was .0674m...the initial angle is 50 degrees from the horizontal. In this problem, I'm trying to find the max height of the grasshoppers jump and the initial velocity of the grasshoppers jump.[/B]


Homework Equations


I've got two relevant equations in my toolbox:

1) Height=)Vo^2(sine of initial theta)^2)/2g


2) Range=s=Vo^2/sin2(theta)*g[/B]


The Attempt at a Solution



I took the range and tried to solve for Vo:

Vo=sqaure root of (1.06m)(9.8m/s^2)(2sin50cos50)=3.2m/s

This was not the correct answer.

I was unable to solve for height given that we currently know only partial height of the grasshoppers trajectory.[/B]

Any help would be greatly appreciated, or advice on how to perceive problems such as these more efficiently. Thanks!
 

Answers and Replies

  • #2
PeroK
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Velocity has two components: horizontal and vertical. Can you see how to split the problem up using this?
 
  • #3
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Well, let's see:

I know that Voy=Vo*sin50 and Vox=Vo*cos50 and Vox=Vx given constant velocity.

Playing around with the constant acceleration equations, I see that:

X=Xo + Voxt+1/2Axt^2 cancels out to X=Vocos50t, then time could be said to be t=1.06m/Vocos50.....

Am I headed in the right direction with this?
 
  • #4
tnich
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X=Xo + Voxt+1/2Axt^2
This doesn't look right. There is no acceleration in the X direction. However, you have come up with a correct equation for t:
t=1.06m/Vocos50
That aside, it is difficult to read your notation. Please use LaTeX.
 
  • #5
PeroK
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Well, let's see:

I know that Voy=Vo*sin50 and Vox=Vo*cos50 and Vox=Vx given constant velocity.

Playing around with the constant acceleration equations, I see that:

X=Xo + Voxt+1/2Axt^2 cancels out to X=Vocos50t, then time could be said to be t=1.06m/Vocos50.....

Am I headed in the right direction with this?

Yes, although perhaps look at the y direction first.
 

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