A stone is thrown vertically upward with a speed of 12m/s from theedge of a cliff 75.0 m high
how much later does it reach the bottom of the cliff?
what is it's speed just before hitting the ground?
what total distance did it travel?
please write initial velocity as u and final one as v
2006-10-01 06:22:22 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
0=75+12*t-.5*9.8*t^2
Using the quadratic formula solve this to get:t=5.32 seconds
Just before hitting the ground the speed is: 12-9.8*5.32=-40.17m/s
To find out how high the stone went, solve .5mv^2=mgh
h=.5v^2/g=7.34 meters. Therefore, the stoned went up 7.34, down 7.34, and then another 75 meters: in total, 7.34+7.34+75=89.69m
2006-10-01 07:36:06 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
kinewho never was good at french lol
2006-10-01 13:28:05 · answer #2 · answered by jackie d 4 · 0⤊ 1⤋