## ksaimouli Group Title 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

1. ksaimouli Group Title

@SithsAndGiggles

2. SithsAndGiggles Group Title

$\text{Range}=\frac{v^2\sin{2\theta}}{g}$ You're given range = 79.5 m v = 30 m/s g = acceleration due to gravity.

3. ksaimouli Group Title

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 Group Title

5. SithsAndGiggles Group Title

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

6. ksaimouli Group Title

yes and this is not the same question i think

7. ksaimouli Group Title

i got 30

8. SithsAndGiggles Group Title

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

9. SithsAndGiggles Group Title

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. Karthik07 Group Title

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

11. Karthik07 Group Title

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