# #325 - EK Physics: Math Error?

Discussion in 'MCAT Study Question Q&A' started by ilovemcat, 01.18.11.

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1. ### ilovemcatBanned

Joined:
04.16.10
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665
Status:
Pre-Medical
I'm have trouble trying to figure out what I'm doing wrong here:

The question states:

A 2 kg ball is thrown upwards with a speed of 40 m/s. At what height will the ball be, when its kinetic energy is equal to its potential energy?

I know that during half the trip (half the height), Potential Energy = Kinetic Energy. In order to solve for height, I use one of the kinematic equations.

But here's where the problem lies, I get two different answers using different equations but can't seem to understand why. They should be equal regardless.

h = hi + vit + .5at^2
h = 0m + (40m/s)(4s) + (5 m/s)(4^2)
h = 240 meters

or using this equation:

vf^2 = vi^2 + 2ah and solving for 2
2gh = vi^2
2(10m/s)h =1600 (m/s)^2
h = 1600 (m/s)^2 / 20
h =80 meters

Any idea what I'm doing wrong to get different answers?

3. ### ilovemcatBanned

Joined:
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By the way, the correct answer is 40 meters. (half of 80 meters)

4. ### ilovemcatBanned

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Ah, nevermind! I just realized I forgot to subtract in the first kinematic equation (160 - 80). I need to rest my brain. I've been working too hard today lol.

5. ### shffl

Joined:
09.20.10
Messages:
130
Remember acceleration is a negative number. If you used -10m/s^2, then your answer would be h = 80m as well

Joined:
09.06.10
Messages:
231
I think the 2nd one saves much more time. At maximum height, final velocity is zero.
vf^2 = vi^2 + 2ax
vf=0
x = vi^2 / 2a
x = 1600 / 20 = 80m so this is the maximum height (all kinetic energy converted back to potential energy)

Halfway height is where KE = PE; 40m.