Question

A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How long is required to reach this height?

For part A) I did: y=11m, g=9.81m/s^2, Vy0= 18 m/s, y0=0, (V^2)y=?

V^2 y= V^2 y0- 2g (y-y0)
--> V^2 y=18^2m/s-2(9.8)(11m-0m)
--> Vy=Square root ( 324m/s-215.6)= 10.4m/s

Part B) I started out with:
y=y0 + Vy0t - (1/2) gt^2
--> 11= 18t m/s-(1/2) 9.8t^2
--> -11 + 18t- 9.8t^2

Answer 1

For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative. 
So Vf = square root [2(-9.8)(11.0) + (18.0)^2] 
Vf = 10.4 m/s your answer is correct. 

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, 11= 18t m/s-(1/2) 9.8t^2 then -11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s