Posted by **yaya** on Monday, March 18, 2013 at 11:41am.

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 -
**Steve**, Monday, March 18, 2013 at 11:50am
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 -
**yaya**, Monday, March 18, 2013 at 11:56am
why the initial velocity u divided 20/√2. i cant understand. can u explain it ..

- biomechanics -
**Steve**, Tuesday, March 19, 2013 at 5:26am
sin 45 = 1/√2

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