ksaimouli
  • ksaimouli
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ksaimouli
  • ksaimouli
@SithsAndGiggles
anonymous
  • anonymous
\[\text{Range}=\frac{v^2\sin{2\theta}}{g}\] You're given range = 79.5 m v = 30 m/s g = acceleration due to gravity.
ksaimouli
  • ksaimouli
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

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ksaimouli
  • ksaimouli
i got that thx what about this
anonymous
  • anonymous
Um, you just asked the same question again. Did you solve for theta?
ksaimouli
  • ksaimouli
yes and this is not the same question i think
ksaimouli
  • ksaimouli
i got 30
anonymous
  • anonymous
Oh, sorry I only read the first half. Just a sec.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Use the formula: Range=( u^2 sin2x )/ g where x is the angle of projection.
anonymous
  • anonymous
u- velocity, g- acc. due to gravity.

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