Ask questions and get helpful responses.

Physics

A baseball is thrown from the roof of 23.5m -tall building with an initial velocity of magnitude 10.0m/s and directed at an angle of 53.1∘ above the horizontal.
a.What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.

b.What is the answer for part (A) if the initial velocity is at an angle of 53.1∘ below the horizontal?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
  1. well, you have

    dx/dt = 10cos53.1° = 6
    dy/dt = 10sin53.1° - 9.8t = 8-9.8t
    y = 23.5+8t-4.9t^2
    y=0 when t=3.15

    At t=3.15,
    dx/dt = 6
    dy/dt = -22.87
    so, the speed is √(6^2+22.87^2) = 23.64 m/s

    Given that, I expect you can do part (b).

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Respond to this Question

First Name

Your Response

Still need help? You can ask a new question.