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

Question

ksaimouli · Answer

@SithsAndGiggles

anonymous · Answer

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

ksaimouli · Answer

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

ksaimouli · Answer

i got that thx what about this

anonymous · Answer

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

ksaimouli · Answer

yes and this is not the same question i think

ksaimouli · Answer

i got 30

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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