Student Doctor Network Forums > Pre-Medical Forums > MCAT Discussions > MCAT Study Question Q&A > #325 - EK Physics: Math Error? PDA View Full Version : #325 - EK Physics: Math Error? ilovemcat01-18-2011, 04:39 PMI'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? ilovemcat01-18-2011, 04:39 PMBy the way, the correct answer is 40 meters. (half of 80 meters) ilovemcat01-18-2011, 04:42 PMAh, 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. shffl01-18-2011, 04:52 PMh = hi + vit + .5at^2 h = 0m + (40m/s)(4s) + (5 m/s)(4^2) h = 240 meters Remember acceleration is a negative number. If you used -10m/s^2, then your answer would be h = 80m as well TwoPaddles01-18-2011, 06:59 PMI 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.