a boy throws a stone straight upward with an initial speed of 15.0m/s. what mazimum height will the stone reach before falling back down.
5 answers
Can you use V^2 = Vo^2 + 2a*distance and set V^2 = 0?
One method to solve this is conservation of energy.
initial kinetic energy = final potential energy
let m = mass of stone
let g = acceleration of gravity
let v = initial velocity
let h = maximum height
(1/2)mv^2 = mgh
So,
v^2 = 2gh
h = (v^2)/g
initial kinetic energy = final potential energy
let m = mass of stone
let g = acceleration of gravity
let v = initial velocity
let h = maximum height
(1/2)mv^2 = mgh
So,
v^2 = 2gh
h = (v^2)/g
23
22.95
final velocity squared = initial velocity squared + 2(acceleration)(displacement)
0 = (15*15) + 2(9.8)x
0 = 225 + 19.6x
-225 = 19.6x
x= 11m
0 = (15*15) + 2(9.8)x
0 = 225 + 19.6x
-225 = 19.6x
x= 11m
0 answers