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 one year 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
 one year 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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SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{Range}=\frac{v^2\sin{2\theta}}{g}\] You're given range = 79.5 m v = 30 m/s g = acceleration due to gravity.

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0A 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

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0i got that thx what about this

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2Um, you just asked the same question again. Did you solve for theta?

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0yes and this is not the same question i think

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2Oh, sorry I only read the first half. Just a sec.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2I think the formula for airtime is something like \[t=\frac{2v\sin\theta}{g}\] I might be wrong; I haven't taken physics in 3 years.

Karthik07
 one year ago
Best ResponseYou've already chosen the best response.0Use the formula: Range=( u^2 sin2x )/ g where x is the angle of projection.

Karthik07
 one year ago
Best ResponseYou've already chosen the best response.0u velocity, g acc. due to gravity.
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