a stone is thrown vertically upward with a speed of 12 m/s from the edge of a cliff 75m high.
1) How much later does it reach the bottom of the cliff?
2) What is its speed just before hitting?
3) What total distance did it travel?
im having so much trouble with this question, u have no idea! im hoping you guys can help me !
thnx for all of your help!
2006-10-05 11:13:25 · 2 answers · asked by Anonymous in Education & Reference ➔ Homework Help
1.0=144-2*9.8*h
h=144/19.6m above the cliff=7.35m
0=12-9.8t
t=12/9.8 till itreaches the maximum height
=1.22s
ht from the ground=75+7.35=82.35m
83.5=1/2*9.8*t^2
t^2=17
t=4.12s
1.time for it to reach the ground=4.12+1.22=5.34s
2.speed before hitting will be 9.8*5.34=52.3m/s
3.totaldistance=75+2(7.35)=89.7m
2006-10-05 11:30:03 · answer #1 · answered by raj 7 · 0⤊ 0⤋
For this question, you need to know the gravity speed of an object, which is 9.8m/s and then go from there...
2006-10-05 18:21:18 · answer #2 · answered by ThornQueen 2 · 0⤊ 0⤋