Answered

IDNLearn.com: Your one-stop destination for reliable answers to diverse questions. Join our Q&A platform to access reliable and detailed answers from experts in various fields.

a ball is thrown straight up from the ground with an unknown velocity. Its reaches its highest point after 3.5s. With what velocity did it leave the ground?

Sagot :

MathPhysidn

Answer:

34.3 m/s

Explanation:

The ball is in free-fall, so it undergoes constant acceleration. The motion can be modeled using kinematic equations, also known as SUVAT equations. The equation we will use is:

v = u + at

where

  • u is the initial velocity
  • v is the final velocity
  • a is the acceleration
  • t is the time

At the maximum height, the final velocity is v = 0 m/s. The time is t = 3.5 s. Taking up to be positive, the acceleration is a = -9.8 m/s². Solving for the initial velocity:

v = u + at

0 m/s = u + (-9.8 m/s²) (3.5 s)

u = 34.3 m/s

The ball was thrown at 34.3 m/s.

Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.