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Finding the horizontal range without initial velocity

  1. May 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A locust jumps at an angle of 43.8 degrees and lands 1.0 m from where it jumped.

    If it jumps with the same initial speed at an angle of 62.4 degrees, how far would it jump then?

    X-component knowns:
    Initial velocity = v(cos(43.8))
    displacement = 1m
    angle = 43.8 degrees

    Y-component knowns:
    Initial Velocity= v(sin(43.8))
    displacement = 0
    acceleration= -9.8m/s^2

    2. Relevant equations
    Equation 1:
    Image78.gif
    Equation 2:
    Image106.gif

    3. The attempt at a solution
    First, I found the distance the projectile traveled at the apex of the y component. I did this by solving
    y=tan(43.2)(0.5m)= .46m. I then made .46m my distance and plugged in the relevant info for equation 2 and solved for initial velocity (in the y component). I got 3.63m/s as my answer. My variable for velocity was (-v^2)(sin(43.2)), so i ended up dividing the other side of the equation by sin(43.2) so that may be where i messed up. I just thought since i was solving for a distance in the y-component, and i was using -9.8m/s^2 as my acceleration that i should be consistent.

    I used my initial velocity in the y-component to solve for time using equation 1. Then I used the solution for time to solve for initial velocity in the x-component using equation 1. Once I solved for that i used the pythagorean theorem to solve for the actual initial velocity.

    To find the range of the x-component with the angel being 62.4 degrees I solved for the y and x components of velocity. I found the time using the y-component of initial velocity and plugging it into the 1st equation. Then used time and the initial velocity in x-component to find the distance using equation 1.

    I got 1.39 meters as my final answer. The answer should be .82 meters according to the given solution.
     
    Last edited: May 12, 2015
  2. jcsd
  3. May 12, 2015 #2

    berkeman

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    Staff: Mentor

    Hi Daniel, and Welcome to the PF.

    I didn't quite follow your technique, and if you sketch the two parabolas, you see that the 2nd distance has to be less than 1m.

    I approached the problem by first writing vx1 and vx2 as functions of vi and the angles given.

    Then write y(t) for each of the two parabolas using your first listed Relevant Equation.

    Then set each y(t) = 0 to solve for whan the bug lands back on the ground, and solve for the two times.

    Then use that information and your equations for vx1 and vx2 to solve for the 2nd distance. Maybe give that approach a try to see how it works for you. Please show us all of your work on it. :smile:
     
  4. May 12, 2015 #3

    gneill

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    Staff: Mentor

    Where did that come from? The trajectory is not a straight line, so no triangle there!

    A small hint for a sneaky method: Have you covered the Range Equation yet?
     
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