A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 18.0 m/s from a height of 4.0 m.
(a) How high does the ball rise from its original position?
m
2007-09-17 15:41:40 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
v^2 = u^2 + 2*g*d
v = initial speed = 18.0m/s
u = final speed = 0
g = -9.8m/s^2 (as the ball is decelerating)
d = unknown
0 = u^2 - 2*g*d
d = (u^2)/(2*g)
= (18*18)/(2*9.8) = 16.53m from the original position
2007-09-17 15:50:35 · answer #1 · answered by Savvy 2 · 0⤊ 0⤋
You can ignore the 4 m since the question asks about distance from the original position.
You can use:
y = (v^2 - v0^2)/(-2g) where v = 0, v0 = 18m/s
Then y = -(18)^2/(-2*9.8) = 16.53 m
2007-09-17 15:51:07 · answer #2 · answered by nyphdinmd 7 · 0⤊ 0⤋
Well,...if you really HATE yourself "THAT" badly.....why not let the ball,....FALL ON YOUR "HEAD" !
2007-09-17 15:51:01 · answer #3 · answered by Anonymous · 0⤊ 2⤋