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Projectile, max height half the range, find angle

  • Thread starter click0420
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,425
46

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.
 
  • #3
2
1
I was able to figure it out but thanks!!
 
  • #4
PeterO
Homework Helper
2,425
46
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)
 

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