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Optimal Throwing Angle

Determine the optimal release angle to maximize distance, assuming no air resistance and a normal Earth-surface gravitational pull using the height of release and the speed of the projectile.

Given a height at which an object is released and its speed, calculate the optimal angle needed to maximize the distance thrown.

Formula

{QuantityVariable["α", "Angle"] == ArcSin[1/(Sqrt[2]*Sqrt[1 + (Quantity[1, "StandardAccelerationOfGravity"]*QuantityVariable["h", "Height"])/QuantityVariable["v", "Speed"]^2])], QuantityVariable["x", "Distance"] == Cos[QuantityVariable["α", "Angle"]]*Quantity[1, "StandardAccelerationOfGravity"^(-1)]*QuantityVariable["v", "Speed"]^2*(Sin[QuantityVariable["α", "Angle"]] + Sqrt[(Quantity[2, "StandardAccelerationOfGravity"]*QuantityVariable["h", "Height"])/QuantityVariable["v", "Speed"]^2 + Sin[QuantityVariable["α", "Angle"]]^2])}

symbol description physical quantity
α release angle relative to horizontal "Angle"
h height of release "Height"
v initial speed "Speed"
x horizontal distance traveled "Distance"

Forms

Examples

Get the resource:

In[1]:=
ResourceObject["Optimal Throwing Angle"]
Out[1]=

Get the formula:

In[2]:=
FormulaData[ResourceObject["Optimal Throwing Angle"]]
Out[2]=

Use some values:

In[3]:=
FormulaData[ ResourceObject[ "Optimal Throwing Angle"], {QuantityVariable["x","Distance"] -> Quantity[10, "Meters"]}]
Out[3]=

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