- #1

jegues

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## Homework Statement

A projectile is launched with an intial speed of 60m/s. If it reaches a maximum height of 150m above this level surface then its horizontal range (in m) is ____ ?

## Homework Equations

v[tex]^{2}[/tex] = vo[tex]^{2}[/tex] + 2ay

y = yo + Vot + 1/2at^2

## The Attempt at a Solution

Vo = 60m/s, then Vox = 60cos(theta); Voy = 60sin(theta) y-yo = 150m

V^2 = Vo^2 + 2ad

0 = 60sin(theta) + 2(-9.8)(150)

theta = 64.6 degrees, so Voy = 54.2m/s

Then,

y-yo = Voyt + 1/2at^2

0 = -150 + 54.2t - 4.9t^2

no zeroes, I was going to grab the zeros for this quadratic and the use, x/2 = Vox/t; x being my final range. Any ideas?

EDIT: I thought about it a bit more and decided that since I already found the angle I could just turn it into a triangle problem.

the triangle has a height of 150, so the distance in the x direction is simply, x = 150/tan(theta);

This will give the HALF of the horizontal range, so 2x will give me my answer. Now oddly enough, 4x seems to give me the correct answer and 2x only half of the correct answer... Coincidence?

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