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### Course: Physics library>Unit 2

Lesson 2: Optimal angle for a projectile

# Optimal angle for a projectile part 4: Finding the optimal angle and distance with a bit of calculus

Created by Sal Khan.

## Want to join the conversation?

• why don't you use formula, that 2cos[a]*sin[a]=sin2[a]?
the formula for distance would be d=(v^2/g)sin2[a]. Wouldn't it be much easier to tell what is the optimal angle for the projectile? You know that highest value of sine is 1, then you put it into sin2a=1 > 2a=90deg > a=45deg.
• This is exactly how my physics teacher proved it so that he wouldn't have to use calculus.
• What is the optimal angle and distance on an incline
• The optimal angle would vary with how steep the incline is. I am pretty certain that the optimal angle will be at least 45 degrees. As the incline gets steeper the optimal angle will get bigger.
• does optimal angle change if you include air resistance?
• Yeah, it definitely would. That's why in golf, javelin, shot put, etc they don't go just firing everything at 45 degrees. You tend to throw things at a lower angle with air resistance.
• I understand the math as to why the optimal angle is 45 degrees, but can someone explain the physics behind it?
• It's about having the projectile traveling as fast as possible, for as long as possible. If the angle is bigger, the projectile will be a lot of time in the air, but traveling really slowly. If the angle is smaller, it will travel faster but fall too soon.

Imagine you want to have a rectangle with the most area you can, but it's height + weight has to be equal to x. How do you draw your rectangle?
• 45 degrees is the optimal angle for max distance, how do we find the angle for max distance when the projectile is launched from a different height than it lands?
• This is a pretty tough question, because the derivation for the max distance is much harder because you don't have a simplified formula for time to work with anymore (since there is an initial value for y0). However, it seems to me that if you have a height that it lands at that is lower than your initial height it is fired at, then having an angle that is lower than 45 degrees would result in the optimal maximum range. If the height that it lands at it higher, then an angle that is higher than 45 degrees would result in the optimal maximum range.
• i do not know calculus and am unsure how Sal derived the equation to 45 degrees?i am also unsure why we used calculus.
• Are there any video's here that show me how to find the optimum angle when launching a projectile above ground level?
• Well, remember that
`x(t) = ( s * cosθ ) * ty(t) = h + ( s * sinθ ) * t - 1/2 * g*t^2`
These are the parametric equations for projectile motion with:
s=initial speed, θ=angle, t=time, g=gravity, and what you are looking for h=initial height.
To find what you are asking for, you have to find the flight time using the quadratic formula on the y equation (to find when y=0) then input that time to the x equation to find the range. Then you can take the derivative ∂/∂θ to find the optimal θ.
• When he plugs in the values for s^2/g why does he write g as 10m/s^2 instead of -10 m/s^2. Is the acceleration not downward?
• You are right! that he should have written -10m/s^2 but you see either way the answer was going to be the same, that is +10m , because you see it is the horizontal "Distance" that he wanted to find, as a function of theta and Distance is a Scaler quantity so no direction involved or no -ve sign, just the magnitude...Either way we would have to neglect that there is a -ve value.
• Excuse my ignorance, however I was curious to exactly what is the Optimal distance mean? Is it the maxima if you were looking at the graph or is the highest point or what? Thanks in advance :)