# Range question

1. Dec 5, 2009

### jegues

1. The problem statement, all variables and given/known data
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 ____ ?

2. Relevant equations

v$$^{2}$$ = vo$$^{2}$$ + 2ay

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

3. The attempt at a solution

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

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?

Last edited: Dec 5, 2009
2. Dec 5, 2009

### PhanthomJay

If you are having trouble solving for t using your quadratic equation, why not use v=gt to solve for t, then double it to get the total flight time. The triangle approach does not work, because the projectile takes on a parabolic shape.

3. Dec 5, 2009

### jegues

Doh, Vfy = Voy -gt1; t1 = Voy/g ; t2 the full time, therefore t2 = 2t1

x = Vox * t2 = 284m,

Thanks.