Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

\[\text{Range}=\frac{v^2\sin{2\theta}}{g}\] 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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
i got 30
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question