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ksaimouli
 2 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
ksaimouli
 2 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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SithsAndGiggles
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i got that thx what about this

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

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

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

SithsAndGiggles
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0u velocity, g acc. due to gravity.
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