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

Do I apply the quadratic formula? I do not know how to answer this, please help! This is due at midnight :(

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

Do I apply the quadratic formula? I do not know how to answer this, please help! This is due at midnight :(

anonymous · Answer

4.9t^2 -18t +11=0
$$\Delta'=81 - 4.9*11 =27.1$$
$$\sqrt{\Delta'} =5.2$$
$$t1 =\frac{ 9-5.2 }{ 4.9 } = 0.77s$$
t2 = 2.9s
you can think an other way
v = -gt +vo
10.4=-9.8t +18
t = 0.77s

masumanwar · Answer

when it reaches maximum height the velocity will be zero and then it will falling under gravitational acceleration

anonymous · Answer

reaches a height=maximum height
oh, sorry, i am poor English