- #1

daniel1991

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## Homework Statement

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

## Homework Equations

Equation 1:

Equation 2:

## 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.

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