2. A golfer hits a golf ball with an initial velocity of 16 m/s at an angle of 44° above the horizontal. The golfer is at an initial height of 15 m above the point where the ball lands. What is the magnitude of the velocity of the ball just before it lands?

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A: launch speed = 8.5 m/sangle = 25 degree above the horizontal

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A:

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A: Ans:- Image-1

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Q: the front and 32 meters fr

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A: a) R = ux×∆t  ux = R/∆t = 92/3.5 = 26.28 m/s ∆t = 2uy/g  uy = ∆tg/2 = (3.5×9.8/2) = 17.15 m/s     b)…

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