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.

Answer 1

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

Answer 2

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

Answer 3

sin 45 = 1/√2