Here's the question you clicked on:
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. At what approximate angle did the ball start? Neglect air drag
\[\text{Range}=\frac{v^2\sin{2\theta}}{g}\] You're given range = 79.5 m v = 30 m/s g = acceleration due to gravity.
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
i got that thx what about this
Um, you just asked the same question again. Did you solve for theta?
yes and this is not the same question i think
Oh, sorry I only read the first half. Just a sec.
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.
Use the formula: Range=( u^2 sin2x )/ g where x is the angle of projection.
u- velocity, g- acc. due to gravity.