Potential and Kinetic Energy with Time Thrown In

[kriegera]

Homework Statement

A stone weighing 0.2 kilograms is thrown vertically upwards with a velocity of 19.6 meters per second. Neglecting the friction of the air, calculate the kinetic and potential energy it possesses at the end of 1 second and 2 seconds respectively.

Homework Equations

The Attempt at a Solution

I found total kinetic energy here:
Total Kinetic Energy=1/2 times mass times velocity squared
KE = (1/2)(0.2)(19.2) ^2 = 36.864
- just not sure how to find it after "1 and 2 seconds respectively."

I know potential energy =
Egrav=mgh

but i don't have a height value so even rearranging doesn't work:
And rearranging to solve for h=Egrav/mg = ?

How do you solve for kinetic and potential energy when you have the added component of time and missing components of height?

 

[rl.bhat]

Use h = ut-1/2gt^2 where u is the initial velocity of the body.
u is given. t is known. Find h at the end of 1 s and 2 s. Then find PE = mgh.
Then KEf = KEi - mgh.

 

[kriegera]

This is what i got:
Use h = ut-1/2gt^2 where u is the initial velocity of the body.
For 1 second h=(19.6)(1)-1/2(9.8)(1) ^2 = 14.7
For 2 seconds h = (19.6)(2) – ½(9.8)(2) ^2 =19.6
PE=mgh
For 1second = PE=(0.2)(9.8)(14.7) = 28.812
For 2seconds = PE (0.2)(9.8)(19.6) = 38.416

For Kinetic Energy-1second
KEf = 36.864 – (0.2)(9.8)(14.7) = 8.052
For Kinetic Energy – 2seconds
Kef = 36.864 – (0.2)(9.8)(19.6) = -1.552

Is this correct? Can that last kinetic energy value be negative?

 

[rl.bhat]

In the second case
final velocity after 2 second is
v = u - gt = 19.6 - 2*9.8 = 0.
So at 2 second KE = 0.

 

[kriegera]

Ok -thanks. Why does the equation change for the second time-for 2 seconds? Why would you use PE-tgh in the first one and H-gt in the second? Is it b/c at 2 seconds, PE would be 0?

 

[kriegera]

one more quick thing - why did we use 0.2 for seconds in the first equation but 2.0 in the second?

 

[rl.bhat]

0.2 kg is mass, and 2 second is the time.