- #1

Vitani11

- 275

- 3

## Homework Statement

A projectile is fired with a velocity v such that it passes through two points both at a distance h above the horizontal. Show that if the gun is adjusted for maximum range, the separation of the points is

d = (v/g)(√v

^{2}-4gh)

## Homework Equations

y = -gt

^{2}/2 + vsinθt

x = vcosθt

for max range R = v

^{2}/g

θ = π/4

## The Attempt at a Solution

substituting theta into the two equations gives me

h = -gt

^{2}/2 + vt/√2

d = (v/√2)t

I replaced y with h to find the two times at which the ball is at h (both going up and down), then I used those times I found in the d equation to find the distance at both points at both times, and then took the difference between "d1" and "d2" where d2 is the point after max height and d1 is the point before. Is this not the right method? I get that t = h and t = stuff from the h equation... and t = h is obviously not true. Is that the issue?