Simple Kinematics Problem, unsure of my error

[N8G]

Homework Statement

Object launched vertically with v initial =20 m/s. No air resistance. How long does it take to reach its max height?

Homework Equations

1/2 at^2 + volt = dy
.5mv^2=mgh

The Attempt at a Solution

I began by using conservation of energy to find the max height. .5mv^2 = mgh
h=(v^2)/2g = 91.74 m
I use that number to solve for t using the quadratic formula.
1/2(-9.81)t^2 + 20t - 91.74 = 0
Using the quadratic formula with this equation gives me imaginary solutions and I'm unsure what error I've made that leads to that.

 

[conscience]

Welcome to PF !

Homework Statement

Object launched vertically with v initial =20 m/s. No air resistance. How long does it take to reach its max height?

Homework Equations

1/2 at^2 + volt = dy
.5mv^2=mgh

The Attempt at a Solution

I began by using conservation of energy to find the max height. .5mv^2 = mgh
h=(v^2)/2g = 91.74 m
I use that number to solve for t using the quadratic formula.
1/2(-9.81)t^2 + 20t - 91.74 = 0
Using the quadratic formula with this equation gives me imaginary solutions and I'm unsure what error I've made that leads to that.

There is a simpler way of doing this problem .

What is the velocity of object on reaching the maximum height ?

Can you think of a kinematic equation relating initial velocity , final velocity and time ?

 

[N8G]

Welcome to PF !
There is a simpler way of doing this problem .

What is the velocity of object on reaching the maximum height ?

Can you think of a kinematic equation relating initial velocity , final velocity and time ?

Yeah, I know that vf = vo +at however that gives me 0 = 20 + (-9.81)t and solving for t gives me a time of 2.039 seconds. However the work that I had already done with energy conservation told me that the max height of the object, that coincides with the point at which its velocity becomes zero, is 91.74 meters. With an initial velocity of 20 m/s, and only traveling for 2.039 sec, the object would never reach 90 meters without accelerating. I just don't understand my error.

 

[TSny]

h=(v^2)/2g = 91.74 m

Check this calculation. If g is in the denominator, then you should use parentheses in the denominator: h=v²/(2g)

 

[conscience]

the max height of the object, that coincides with the point at which its velocity becomes zero, is 91.74 meters.

Check your calculations .

 

[N8G]

(v^2)/(2g) = 400/(2*9.81) = approximately 20

Okay awesome. I must've made an egregious calculator error in the beginning. Thank you!