posted by yaya .
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.
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
why the initial velocity u divided 20/√2. i cant understand. can u explain it ..
sin 45 = 1/√2