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To solve this problem you use the derived formula ∆dh=(V1h2(sin2Ѳ))/g 250m=V1h2(sin2(17))/9.8m/s2 V1h=((sin(17))/9.8m/s2)1/2 V1h= 64.7292201m/s V1h= 65m/s

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Q: A golfer strikes a golf ball at an angle 17 degree above the horizontal. with what velocity must the ball be struck in order to reach the green which is a horizontal distance of 250m from the golfer.?
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A golfer strikes a golf ball at an angle 17 degree above the horizontal. with what velocity must the ball be struck in order to reach the green which is a horizontal distance of 250m from the golfer?

To solve this problem you use the derived formula ∆dh=(V1h2(sin2Ѳ))/g 250m=V1h2(sin2(17))/9.8m/s2 V1h=((sin(17))/9.8m/s2)1/2 V1h= 64.7292201m/s V1h= 65m/s


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What travels farther and arrow shot at a 30 degree angle or 60 degree angle?

Both the arrows will cover equal horizontal distance as their angles are at an difference from 45 degree. Horizontal range is maximum at 45 degrees and decreases equally on sum or difference of an angle from 45. But vertical distance increases on addition of an angle from 45 and decreases on subtraction of angle from it. For more details, contact at saqibahmad81@yahoo.com


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How do you make an angle with horizontal direction?

90 degree