A baseball of mass .5 kg is thrown directly upward from height 0 with an initial

Question

A baseball of mass .5 kg is thrown directly upward from height 0 with an initial
velocity of 50m/sec. Assume that the force in newtons due to air resistance is ?.3v,
where v is the velocity of the object. Determine the equation of motion of the baseball.
What is the maximum height that the baseball reaches? How does your answer
compare to what the maximum height would be if there is no air resistance? Warning:
if you set up coordinates so that the positive axis is in the up direction, then the force
due to gravity is downward, hence negative

Explanation / Answer

The ball will rise until the negative pull causes it to reverse and fall. So the time is -16t^2 + 50t = 0 16t = 50 and the time is 50/16 = 3.125 sec. The height will be the time rising times the velocity plus the initial 6' so its 50 x 3.125 +6 = 162.25' The equation for the second part is -16t^2 +70t +5 so the time is 70/16 sec and the height is 311.25'

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