Question

A stone is thrown vertically upward with a speed of 18.0 m/s. How fast is it moving when it reaches a height of 11.0m? How long does it take to reach the height?

Answer 1

Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2 

v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared 

vf^2=v0^2 + (2)(g)(yf-y0) 
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m) 
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2) 
vf^2=108.18 m^2/s^2 
vf=sqrt(108.18 m^2/s^2) 
vf=10.4 m/s