Astudent wish to measure the gravitational acceleration g. she does it by releasing a small lead ball from rest and measures the time t it takes the ball to drop a distance s. g can then be found using the relation t2. she measures t using an automated digital timer which manual states that it measures drop times up to 500 ms with an uncertainty of 3 ms. she sets the system up such that s 100.0 t 0.3 mm (measured using a measurement tape), and the timer reads that the drop time is 144 1 ms. calculate her best estimate of g and its uncertainty . discuss how she could set up her experiment such that she would achieve a better precision in a single drop, without needing a more precise timer or being able to measure distances with higher precision

(9.64 +- 0.86) m/s^2

Explanation:

The generic motion equation for constant acceleration is

Where

X0: initial position

v0: initial speed

a: acceleration

t: time

If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:

And the acceleration can be obtained as:

Where x is the distance fallen and a = g.

So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate

For the uncertainty we have to calculate the relative uncertainties first

For the distance (100 * 0.3)/100 = 0.3%

For the time (100 * 3)/144 = 2.08%

For multiplications or divisions the relative uncertainties are added

0.3% + 2.08% + 2.08% = 4.46%

We convert this into absolute uncertainty:

(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)

Finally, this is multiplied by a constant scalar, so:

2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)

We convert the units

0.86 m/(s^2)

And the measurement is (9.64 +- 0.86) m/s^2

A better method is putting the ball in a ramp instead of a free fall, that way the fall is longer and the effect of time measuring uncertainty is reduced.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s