- #1

click0420

- 2

- 1

## Homework Statement

find the angle of the projection for which the maximum height is equal to half of the range

## Homework Equations

vertical: h =(vsinθ)t-0.5(g)(t^2)

horizontal: 2h =(vcosθ)t

## The Attempt at a Solution

i know you set both equations equal to each other

(vsinθ)t -(0.5)(g)(t^2) = ((t)vcosθ)/2

cancel t

(vsinθ)-(0.5)(g)(t) = (vcosθ)/2

(2vsinθ-vcosθ)/2 = (0.5)(g)(t)

at this point i am slightly stuck, i know i need to sub some equivalent form in for "t" though i am not sure how/what formulas to use.