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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