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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

@SithsAndGiggles

2. 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.

3. 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

4. ksaimouli

i got that thx what about this

5. anonymous

Um, you just asked the same question again. Did you solve for theta?

6. ksaimouli

yes and this is not the same question i think

7. ksaimouli

i got 30

8. anonymous

Oh, sorry I only read the first half. Just a sec.

9. 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.

10. anonymous

Use the formula: Range=( u^2 sin2x )/ g where x is the angle of projection.

11. anonymous

u- velocity, g- acc. due to gravity.

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