Asked by Anonymous
A golfer wants to drive a ball a distance of 310 yards. If the 4-‐wood launches the ball at an angle at 15° above the horizontal, and assuming that the drive is over level ground (i.e. the start and final vertical position is the same; vertical displacement is zero), what must be the initial speed of the ball to achieve the required distance (assume a perfectly straight drive).
Answers
Answered by
bobpursley
Hmmm. The ball does not roll or bounce?
Time in air:
in the vertical, hf=hi+Vsin15*t-1/2 g t^2
or 0=(VSin15-4.9t)t solve for t.
then horizontal distance..
310yards(39.3inches/yard)(.0254m/inch)=Vcos15*t
solve for V
Time in air:
in the vertical, hf=hi+Vsin15*t-1/2 g t^2
or 0=(VSin15-4.9t)t solve for t.
then horizontal distance..
310yards(39.3inches/yard)(.0254m/inch)=Vcos15*t
solve for V
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