<p>Ok my teacher gave me this problem and I am not sure hot to do it. I think I get more confused w/ the naming etc.</p>
<p>A ball is thrown vertically upward with a velocity of magnitude (C1) from a window at height H. What is the ball's position as a function of time? How high will it go before starting to fall? When will it hit the ground?</p>
<p>Thanks...I just have no clue where to start or anything!</p>
<p>It's a parabolic function the ball is thrown up with an initial velocity of C1 it decelerates to 0 at it's hightest point and then will reach C1 velocity at the same point it started. Then it will continue to accelerate till it hits the ground. If it is thrown vertically upward, it has no horizontal component so all the velocity is in the y direction.</p>
<p>The ball is decelerating at 9.8 m/s2 so:
Vfinal = Vinitial-gt or vfinal = C1 - 9.8t
ball's position as a funtion of time y - yinitial = vinitialt-1/2gt^2 or
y = C1t - 1/2(9.8)t^2 + H</p>
<p>Max height is when vfinal = 0 so 0 = C1-9.8t or t = C1/9.8 then plug t into y = C1 - 1/2 (9.8)t^2 + H to get it's height above the ground</p>
<p>It will hit the ground when y = 0 so
0 = C1t - 1/2(9.8)t^2 + H and solve for t</p>