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:
[tex]v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}[/tex]
Similarly, The equation of mption:
[tex]x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})[/tex]
replacing our assumed values:
where [tex]v_=0 \ and \ g= 9.81[/tex]
[tex]x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})[/tex]
[tex]x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m[/tex]
[tex]\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}[/tex]
So, when the object hits the ground when x(t) = 1000
Then from above derived equation:
[tex]\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}[/tex]
[tex]1000= 0.981t-0.981(1-e^{-(10)t}) \ m[/tex]
By diregarding [tex]e^{-(10)t} \ m[/tex]
[tex]1000= 0.981t-0.981[/tex]
1000 + 0.981 = 0.981 t
1000.981 = 0.981 t
t = 1000.981/0.981
t = 1020.36 sec
