Question

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant b ​N-sec/m, determine the equation of motion of the object. When will the object strike the​ ground? Assume that the acceleration due to gravity is and let​ x(t) represent the distance the object has fallen in t second

Answer 1

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec