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A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR. a) At what angleθis therock thrown? b) In terms of its original rangeR, what is the rangeRmaxtherock can attain if it is launched at the same speed but at the optimal anglefor maximum range? c) Would your answer to part a) be different if the rockis thrown with the same speed on a different planet? Explain.

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a) The vertical component of velocity v is taking the rock to a height

Vertical component = [tex]vsin\theta[/tex]
The time taken to reach maximum height = [tex]\frac{vsin\theta}{g}[/tex]
So total time of rocks flight = [tex]\frac{2vsin\theta}{g}[/tex]
Range of rock is due to the horizontal component of velocity = [tex]vcos\theta[/tex]
Range = [tex]\frac{2*v*sin\theta*v*cos\theta}{g}[/tex] = [tex]\frac{2*v^2*sin\theta*cos\theta}{g}[/tex]
Maximum height = [tex]\frac{g*t^2}{2}[/tex] = [tex]\frac{v^2*sin^2\theta}{2*g}[/tex]
Since range = maximum height
We have [tex]\frac{2*v^2sin\theta*cos\theta}{g} = \frac{v^2*sin^2\theta}{2*g}[/tex]
[tex]tan\theta = 4[/tex]
[tex]\theta = 75.96^0[/tex]
So when angle of projection is [tex]\theta = 75.96^0[/tex] range is equal to maximum height reached.
b) We have range = [tex]\frac{2*v^2*sin\theta*cos\theta}{g}[/tex] =[tex]\frac{2*v^2*sin2\theta}{g}[/tex]
Maximum of range is reached when [tex]\theta = 45^0[/tex]
Maximum range = [tex]\frac{2*v^2}{g}[/tex]
c) For range to be equal to maximum height only condition is [tex]tan\theta = 4[/tex], it does not depend upon acceleration due to gravity and velocity. That angle is a constant.

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