Post a New Question


posted by .

A Softball is Thrown at an angle of 45 degrees with the horizontal. It leaves the hand 2m above the ground. Its Initial Velocity is 20 m/s . Neglecting all external resistance, how far would it travel ? The point at which it hits the ground is at the same level as the ground at the point from which it was thrown.

  • biomechanics -

    The vertical component of the initial velocity is 20/√2 = 14.14 m/s

    So, the height

    y = 2 + 14.14t - 4.9t^2
    when does y=0?
    At t=3.02 sec

    Since the horizontal component of the velocity does not change, in those 3.02 seconds, the ball traveled

    3.02*14.14 = 42.7 m

  • biomechanics -

    why the initial velocity u divided 20/√2. i cant understand. can u explain it ..

  • biomechanics -

    sin 45 = 1/√2

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question