A Princeton prof. (mass = 70.0 kg), surprised by the large stopping force he calculates for jumping flat footed from a height of 0.12 m, decides to try the experiment. Calculate he deceleration (in g's) if he stops in a distance of 0.36 cm. (Do not try this. You could easily break an ankle!)
V at impact = sqrt (2 g H) = 1.53 m/s
If a is the deceleration after impact,
V = sqrt (2 a d), where d = 3.6*10^-3 m
a = V^2/(2 d) = 327 m/s^2
(33 g's)
Why did the Princeton professor try such a dangerous experiment? Did he lose a bet with Newton or something? Anyway, let me calculate the deceleration for you.
First, let's convert the stopping distance to meters:
0.36 cm = 0.36/100 = 0.0036 m.
Now, we can use the formula for deceleration:
deceleration = (final velocity^2 - initial velocity^2) / (2 * stopping distance).
Assuming the professor starts from rest (initial velocity = 0 m/s), the final velocity can be calculated using the equation:
final velocity^2 = 2 * acceleration * distance.
Substituting the values:
final velocity^2 = 2 * acceleration * 0.12 m.
Now, let's solve for final velocity:
final velocity^2 = 0.24 * acceleration.
Substituting the values for stopping distance and solving for acceleration:
deceleration = (0 - 0)^2 / (2 * 0.0036 m) = 0 / 0.0072 m = 0 m/s^2.
Hmm, that's strange. It seems like the professor experienced no deceleration at all! Maybe he should try a different experiment like figuring out how many jokes it takes to confuse a physicist.
To calculate the deceleration experienced by the professor, we can use the equation:
deceleration = (Vf^2 - Vi^2) / (2 * d)
where Vf is the final velocity, Vi is the initial velocity (which is 0 as the professor jumps flat footed), and d is the distance over which the professor stops.
Given:
Mass of the professor (m) = 70.0 kg
Height (h) = 0.12 m
Stopping Distance (d) = 0.36 cm = 0.0036 m
First, let's calculate the initial velocity (Vi) using the equation:
Vi = √(2gh)
where g is the acceleration due to gravity (9.8 m/s^2).
Vi = √(2 * 9.8 * 0.12) = √(2.352) ≈ 1.535 m/s
Next, let's calculate the final velocity (Vf) using the equation:
Vf = sqrt(Vi^2 + 2ad)
where a is the acceleration (deceleration) and d is the stopping distance.
Vf = sqrt(1.535^2 + 2 * a * 0.0036)
Since the professor comes to a stop, Vf = 0:
0 = √(1.535^2 + 2 * a * 0.0036)
Squaring both sides and solving for a, we get:
a = - Vi^2 / (2 * d)
a = - (1.535^2) / (2 * 0.0036)
a ≈ -334.52 m/s^2
To convert this acceleration to g's, we divide by the acceleration due to gravity:
a_g = a / g
a_g ≈ -334.52 m/s^2 / 9.8 m/s^2
a_g ≈ -34.14 g's
Therefore, the deceleration experienced by the professor when stopping in a distance of 0.36 cm is approximately -34.14 g's.
To calculate the deceleration (in g's) experienced by the Princeton professor, we need to use the following formula:
a = (v^2 - u^2) / (2s)
Where:
a = acceleration
v = final velocity
u = initial velocity
s = stopping distance
In this case, the professor jumps flat-footed from a height of 0.12 m and stops in a distance of 0.36 cm. The initial velocity (u) is 0 as the professor starts from rest.
First, let's calculate the final velocity (v):
Using the equation of motion: v^2 = u^2 + 2as
Since u = 0, the equation becomes: v^2 = 2as
Plugging in the values: v^2 = 2 * 9.8 * 0.12
Simplifying: v^2 = 2.352
Taking the square root of both sides: v ≈ 1.533 m/s
Now, we can calculate the deceleration (a) in m/s^2:
a = (v^2 - u^2) / (2s)
= (1.533^2 - 0^2) / (2 * 0.0036)
= 2.352 / 0.0072
≈ 326 m/s^2
Lastly, let's convert this acceleration into g's by dividing it by the acceleration due to gravity:
g = 9.8 m/s^2
Deceleration (in g's) = 326 / 9.8
≈ 33.27 g's
Therefore, the deceleration experienced by the Princeton professor when stopping in a distance of 0.36 cm is approximately 33.27 g's.