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ksaimouli

  • 3 years ago

A golfer hits the ball with a speed of 30 m/s. The ball touches the ground on the green 79.5 m away. At what approximate angle did the ball start? Neglect air drag

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  1. ksaimouli
    • 3 years ago
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    @SithsAndGiggles

  2. SithsAndGiggles
    • 3 years ago
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    \[\text{Range}=\frac{v^2\sin{2\theta}}{g}\] You're given range = 79.5 m v = 30 m/s g = acceleration due to gravity.

  3. ksaimouli
    • 3 years ago
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    A golfer hits the ball with a speed of 30 m/s. The ball touches the ground on the green 79.5 m away. Approximately, how long was the ball in the air? Neglect air drag

  4. ksaimouli
    • 3 years ago
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    i got that thx what about this

  5. SithsAndGiggles
    • 3 years ago
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    Um, you just asked the same question again. Did you solve for theta?

  6. ksaimouli
    • 3 years ago
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    yes and this is not the same question i think

  7. ksaimouli
    • 3 years ago
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    i got 30

  8. SithsAndGiggles
    • 3 years ago
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    Oh, sorry I only read the first half. Just a sec.

  9. SithsAndGiggles
    • 3 years ago
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    I think the formula for air-time is something like \[t=\frac{2v\sin\theta}{g}\] I might be wrong; I haven't taken physics in 3 years.

  10. Karthik07
    • 3 years ago
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    Use the formula: Range=( u^2 sin2x )/ g where x is the angle of projection.

  11. Karthik07
    • 3 years ago
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    u- velocity, g- acc. due to gravity.

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