Question

An object of mass 10kg is released from rest 1000m 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 bequals40​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 9.81 m divided by sec squaredand let​ x(t) represent the distance the object has fallen in t seconds.

Answer 1

Answer:

x(t) =  2.4525 t -0.6131 × ( 1 - $e^{4t}$ )

t = 408 second

Explanation:

given data

mass m = 10 kg

height x(t)  = 1000 m

b = 40 N-s/m

acceleration due to gravity g = 9.81 m/s²

solution

we know here that equation of motion here v(t) is express as

$v(t) = \frac{mg}{b} + (vo - \frac{mg}{b}^{e^{bt/m}} )$          ......................1

and

x(t) will be express here as

$x(t) = \frac{mg}{b} t + \frac{m}{b} (vo -\frac{mg}{b})\times ( 1- e^{bt/m})$    .......................2

now put here value in equation 2 and we will get

x(t) = $\frac{10\times 9.81}{40} t+ \frac{10}{40} (0 - \frac{10 \times 9.81}{40}) \times ( 1- e^{40t/10})$    

x(t) =  2.4525 t -0.6131 × ( 1 - $e^{4t}$ )

and

now we get here time t after object hit 1000 m by height

put here value x(t) we get t

1000 =  2.4525 t -0.6131 × ( 1 - $e^{4t}$ )

solve it and we get by neglecting $e^{4t}$

t = 408 second