Projectile, max height half the range, find angle

[click0420]

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.

 

[PeterO]

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.

Go for concepts here.

At maximum height, the object is half way (horizontally) to its landing point - so at that point the height is 1/4 of the range.

The vertical part of the trip has been done while slowing down, so the average speed is 1/2 the initial speed.

It gets to a height 4 times its range so far, so the average vertical speed is 4x the horizontal speed.

Thus the initial vertical speed is 8x the horizontal speed.

Apply Pythagoras to that.

 

[click0420]

I was able to figure it out but thanks!

 

[PeterO]

I was able to figure it out but thanks!

Re-reading to see that you were only after the angle, you merely wanted tan^-1(8)