vertical acceleration

Asperas, Marco Physics 11 Mr. Cartajenas Experiment September 5, 2006 Vetical Acceleration Problem: Vertical acceleration is a constant value. This investigation will use that concept to calculate the vertical height of the CIS building. Hypothesis: 6.096m (10ft/storey, 10ft ≈ 3.048m, 3.048 * 2 = 6.096) Materials: 1. Ball 2. Stopwatch Procedure: 1. Position one member at the top of the building and one at the bottom. The member on top holds the ball and the one at the bottom stands close to the landing site with the stopwatch. 2. Drop the ball and immediately start timing. 3. Obtain the time it takes for the ball to drop (in seconds). 4. Repeat the process four times and get the average time. 5. Using the formula y = y0 + ½ gt^2 (g = -9.8m^2), find the distance (y, in meters) of the building. Data: Trial Time (seconds)( ±0.06) 1 1.13 2 1.21 3 1.15 4 1.17

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Newton's Laws of Motion. Second law of motion: Law of acceleration. Acceleration affected by the force of gravity.

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Asperas, MarcoPhysics 11Mr. CartajenasExperimentSeptember 5, 2006Vetical AccelerationProblem: Vertica acceeration is a constant !a"e.#his in!esti$ation %i "se that concept to cac"ate the !ertica hei$ht o& the C'S b"i(in$.Hypothesis: 6.0)6m *10&t+storey, 10&t , -.0./m, -.0./ 0 2 1 6.0)62Materials:1. 3a2. Stop%atchProcedure:1. Position one member at the top o& the b"i(in$ an( one at the bottom.#hemember on top ho(s the ba an( the one at the bottom stan(s cose to the an(in$ site %ith the stop%atch.2. 4rop the ba an( imme(iatey start timin$.-. 5btain the time it ta6es &or the ba to (rop *in secon(s2... 7epeat the process &o"r times an( $et the a!era$e time.5. 8sin$ the &orm"a y 1 y0 9 : $t;2 *$ 1

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