Question

1. [25 points) Idealized frictionless free fall of an object that is dropped from being at rest at i = 0. For the following question, to model the free fall of a falling rock, assume the usual idealizing simplifications for solving "free fall" problems. Consider the following experiment. A rock with a mass of m= 2 kg is dropped at the time t = 0 from a height of 140 m above ground. Assume that the rock is simply dropped at time t = 0 s, without it having any initial velocity. Let s(t) denote the function that describes the height of the rock above ground during its free fall, and let v(t) denote its velocity, and a(t) denote its acceleration at any given time to (a) Sketch a drawing of this experiment. (b) Using the relevant facts from physics and calculus, state the second order differential equation that describes the function s(t). Furthermore, state the initial conditions on s(t) that describe the rock's initial height and the fact that it is dropped from being at rest. (e) Find the general solution s(t) of the differential equation in b). (d) Find the solution s(t) of the initial value problem in b). (e) Use your answer in d) to find the exact time that when the rock hits the ground, and the velocity hit with which the rock hits the ground. (f) Assuming that the rock comes to an immediate rest when it hits the ground, sketch? - the graph of the position function s(t), - the graph of the velocity function r(t), and - the graph of the acceleration function alt) over the time interval (0.7). Hint. This question is very similar to Example 1 on page 2 of the handwritten notes of class 27. That is, assume that the Earth's gravitational acceleration g is constant throughout the fall, with g = 9.8 m/s, and assume furthermore that no force other than the gravitational force F, is acting on the rock during the fall; for instance, the usual force due to air resistance is ignored. For more details on this, see class 27 and its handwritten notes "We discussed all of these facts in class 26; see for instance pages 2, 8, 9 and 10 of its handwritten notes If you work correctly, you will see that th5.345 s. That is, assume that it does not bounce at all once it hits the ground. Use the horizontal axis to measure the time t, and the vertical axis for the particular function in question. The time interval (0.7) is the interval containing all the possible times between t = 0s and 1 = 7s.

Answer 1

Rate of change of velocity is acceleration, as in free falling of mass. Acceleration is gravitational acceleration that g = 9.8m/s^2

thus,

$a(t)=\frac{d^2s}{dt^2}=\frac{dv}{dt}=-9.8$m/s^2

acceleration is taken negative because it is acting downward.

integrating the differential equation:

$\\\int_{0}^{v}dv=-\int_{0}^{t}9.8dt\\\\ v=-9.8t$m/s

Rate in change in displacement is velocity:

$\frac{ds}{dt}=-9.8t$

Integrating this:

(this is function of displacement from initial position with respect to time)

Since ds 1 9.8tdt 140 +2 S – 140 = -9.82 2 s = 140 - 4.912

when rock hit the ground:

s = 0

$140=4.9t^2;\: \: \Rightarrow t_{hit}=\sqrt{140/4.9}\approx 5.35$second

velocity for the same:

$v_{hit}=-9.8\sqrt{140/4.9}\approx -9.8*5.35=-52.38$ m/s

(negative because direction of velocity is downward)

Position function graph. Impact at t = 5.35second. then velocity is zero shown till t = 7 second.

(quadratic function before impact), no motion after impact

s(t) 150 s(t)=140-4.9t^2 100 50 0 2 6 7 7 8 10 t

Velocity function graph: (linear function)

Velocity become zero after impact

0 2 3 4 5 6 7 -10 --20 --30 --40 -50 --60

Acceleration function graph (constant function)

Acceleration become zero after impact

$1590519052668_image.png$