a rock is thrown straight up with an initial velocity of 20 m/s. (How high) does the ball rise when it reaches its ( Highest point)?
a = graivty = - 9.8 m/S^2 (that lil 2 ontop of the S) when objects are free falling
(How high) - pf = ?
(highest point) - vf= 0 m/s
2006-09-17 17:18:01 · 4 answers · asked by sub.lihhjj 2 in Science & Mathematics ➔ Physics
It takes 20/9.8=2.04 seconds for the velocity to equal zero. In that time, the ball would travel x= 20*2.04 -.5*9.8*(2.04)^2=20.4 meters.
2006-09-17 17:28:24 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
V^2 = 2 A S Where v=Velocity by which stone is thrown, A= Acceleation due to gravity, S=Distance covered
Hence;
20*20/(9.8*2) = Desired Answer (S)
Therefore S=20.40 meter
2006-09-17 18:37:56 · answer #2 · answered by Mihir Durve 3 · 1⤊ 0⤋
The height, when the rock reaches its highest point is given by
h = (u^2) / 2g
h= height attained by the rock
u=initial velocity
g=gravity(9.8 m/s^2)
h= (20 * 20)/(2 * 9.8) mts.
its 20.40 mts
2006-09-17 17:34:40 · answer #3 · answered by testy c 1 · 1⤊ 0⤋
very high.
2006-09-17 17:37:20 · answer #4 · answered by jimmy 1 · 0⤊ 1⤋