Max horizontal distance wrt variable angle

In summary: So the maximum distance the ball can reach is when it is thrown with an angle of 2h+v^{2}sin(\theta).
  • #1

Homework Statement


A ball is thrown with speed v from height h. What angle should it be thrown so it travels the maximum horizontal distance?

I'll assume the angle theta is the angle from the horizontal to the initial direction of the velocity.

Okay, since there is no air resistance we know:
[tex]x = x_{0} + v_{0x}t + \frac{1}{2}a_{x}t^{2}[/tex]
since there is no acceleration in that direction, the distance the ball can reach would be:
[tex]x = v_{0}cos(\theta)t[/tex]

Since, as it stands, x is a function of both t and theta I will use the vertical motion to find the time of flight as a function of theta and plug that into the above equation.

I think I have to do this in two parts (or maybe I've been working too hard and have confused myself).

Part1: from launch to maximum height (ie v_y = 0)
Part2: from maximum height to landing

Part1:
[tex]y = h + v_{0}sin(\theta)t - \frac{1}{2}gt^{2}[/tex]
We don't know what the maximum height is but we can take the derivative to find at what time it occurs (and thus find it).
[tex]v_{y} = v_{0}sin(\theta) - gt[/tex]
[tex]t_{1} = \frac{v_{0}sin(\theta)}{g}[/tex]

Plugging this into the first equation of Part 1 yields:
[tex]y_{max} = h + \frac{1}{2g}v_{0}^{2}sin(\theta)^{2}[/tex]

Part2:
[tex]y = y_{max} + v_{0}t - \frac{1}{2}gt^{2}[/tex]
here v_0 is zero and y_final = 0
[tex]y_{max} = \frac{1}{2}gt^{2}[/tex]
after some simplifying:
[tex]t_{2} = \frac{1}{g}\sqrt{2gh + v_{0}sin(\theta)^{2}}[/tex]

So the total time of flight would be:
[tex]t = t_{1} + t_{2} = \frac{v_{0}sin(\theta) + \sqrt{2gh + v_{0}^{2}sin(\theta)^{2}}}{g}[/tex]

plugging this into the equation for horizontal distance:
[tex]x = v_{0}cos(\theta)(\frac{v_{0}sin(\theta) + \sqrt{2gh + v_{0}^{2}sin(\theta)^{2}}}{g})[/tex]

now, as a check, letting h = 0 we would get:
[tex]x = \frac{2v_{0}^{2}}{g}cos(\theta)sin(\theta)[/tex]
which gives an angle of 45 degrees being optimal (as kinda expected).

Now, without that simplification that above expression is nasty and I can't find a way to simplify it or its derivative which leads me to believe I messed up earlier >.<
 
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  • #2
hi alexa! :smile:

(have a theta: θ and a square-root: √ and try using the X2 and X2 icons just above the Reply box :wink:)
iamalexalright said:
I think I have to do this in two parts (or maybe I've been working too hard and have confused myself).

Part1: from launch to maximum height (ie v_y = 0)
Part2: from maximum height to landing

sorry, but yes you've been working too hard …

get some sleep! :zzz: …

then just go straight to solving the original equation to find t when y = 0 :wink:
 
  • #3
Ha... maybe I could use a nap...

But I recover the same travel time (as I should) and I'm still left with something that I can't simplify.
 
  • #4
show us your solution for y = 0, so that we can see what's going wrong :wink:
 
  • #5
okay, I start here:
[tex]y = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}[/tex]
we know y = 0
y_0 = h
v_0y = v_0 sinθ
[tex]0 = h + v_{0}sin(\theta)t - \frac{1}{2}gt^{2}[/tex]
I'll use the quadratic formula to find the roots:

[tex]t = \frac{1}{g}(v_{0}sin(\theta) + \sqrt(v_{0}^{2}sin{2}(\theta) + 2gh)[/tex]
 
  • #6
ok, now multiply by cosθ and find the maximum :wink:
 
  • #7
well that's my issue! To find the maximum I'd have to take a derivative wrt theta, no?

[tex]x = \frac{1}{g}(v_{0}^{2}sin(\theta)cos(\theta) + v_{0}cos(\theta)\sqrt{2gh + v_{0}^{2}sin^{2}(\theta)}[/tex]
 
  • #8
yes! :rolleyes:

do it!
 
  • #9
[tex]\frac{v_{0}^{2}}{g}cos(2x) + \frac{vsin(\theta)(v^{2}cos(2\theta) - 2gh)}{g\sqrt{2gh + v^{2}sin^{2}(\theta)}}[/tex]

Setting it equal to zero and doing some simplifying:

[tex]cos(2\theta)\sqrt{2gh + v^{2}sin^{2}(\theta)} = 2ghsin(\theta) - v^{2}cos(2\theta)sin(\theta)[/tex]

more simplifying doesn't seem to get me anywhere... although I just missed a sin in there, let me see what I can do now...
 

What is the concept of "Max horizontal distance wrt variable angle"?

Max horizontal distance wrt variable angle refers to the maximum distance that an object can travel horizontally when launched at different angles. This concept is often used in physics and engineering to determine the optimal angle for launching projectiles, such as rockets or projectiles.

How is "Max horizontal distance wrt variable angle" related to projectile motion?

Max horizontal distance wrt variable angle is directly related to projectile motion because it involves the horizontal displacement of an object that is launched at different angles. Projectile motion is a type of motion in which an object is launched into the air and moves along a curved path under the influence of gravity.

What factors affect the "Max horizontal distance wrt variable angle"?

The factors that affect the max horizontal distance wrt variable angle include the initial velocity of the object, the angle at which it is launched, and the force of gravity. Other factors, such as air resistance and wind direction, may also play a role in the maximum distance achieved.

How can the "Max horizontal distance wrt variable angle" be calculated?

The max horizontal distance wrt variable angle can be calculated using the equations of projectile motion, which take into account the initial velocity, launch angle, and acceleration due to gravity. Alternatively, it can be determined experimentally by measuring the distance traveled by an object at different launch angles.

What is the practical application of understanding "Max horizontal distance wrt variable angle"?

Understanding the concept of max horizontal distance wrt variable angle is crucial in many fields, including physics, engineering, and sports. It allows engineers to design optimal trajectories for projectiles, and it helps athletes determine the best angle for launching objects, such as in the sport of javelin throwing.

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