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)

 

[Nickie]

Hello! I would approach this problem by first identifying the variables and their relationships in the equations provided. The variables in this problem are the angle of projection (θ), initial velocity (v), time (t), and acceleration due to gravity (g). The equations provided represent the vertical and horizontal components of projectile motion, which are related by the time of flight (t).

To find the angle of projection for which the maximum height is equal to half of the range, we can set the equations equal to each other and solve for θ. However, as you mentioned, we need to find an equivalent form for t in order to do this.

One way to approach this is to use the horizontal equation to solve for t in terms of θ. We can rearrange the equation to get t = (2h)/(vcosθ). Then, we can plug this into the vertical equation to get h = (v^2sin^2θ)/(2gcos^2θ).

Now, we can set this equal to half of the range (which we can express as (v^2sin2θ)/g) and solve for θ. This will give us the angle of projection for which the maximum height is equal to half of the range.

I hope this helps! Keep in mind that there may be other approaches to solving this problem, so don't get discouraged if you're not sure about the method you've chosen. As a scientist, it's important to be open to different ways of thinking and problem solving. Good luck!