Post a New Question

physics

posted by on .

A ball is thrown straight upward and returns to the thrower's hand after 2.20 s in the air. A second ball is thrown at an angle of 31.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?

  • physics - ,

    You need the vertical velocities the same.

    time in air for the second ball.
    vertical velocity at the top is zero.

    Vf=Vsin31-g*t
    Vf=0, so t=Vsin31/g but this time is 1.1 seconds, so solve for V to make that happen.

  • physics - ,

    26.7081046958

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question