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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. At what approximate angle did the ball start? Neglect air drag
 one year ago
 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
 one year ago

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SithsAndGigglesBest 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.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
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
 one year ago

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

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

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

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

SithsAndGigglesBest ResponseYou've already chosen the best response.2
I 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.
 one year ago

Karthik07Best ResponseYou've already chosen the best response.0
Use the formula: Range=( u^2 sin2x )/ g where x is the angle of projection.
 one year ago

Karthik07Best ResponseYou've already chosen the best response.0
u velocity, g acc. due to gravity.
 one year ago
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