Chapter 3: Q6E (page 115)

An object of mass 8 kg is given an upward initial velocity of 20 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -16v , where v is the velocity of the object in m/sec. Determine the equation of motion of the object. If the object is initially 100 m above the ground, determine when the object will strike the ground.

Short Answer

Expert verified

The equation of motion of an object $x (t) = 4.9 t - 12.45 e^{- 2 t} + 112.45$ and the time takes by the object to strike the ground is 22.9 sec.

Step by step solution

Important hint.

Use Newton’s method to solve for t.

$t_{n + 1} = t_{n} - \frac{f (t_{n})}{f' (t_{n})}$

Find the velocity

For finding the weight of the object apply:

Net force=W-drag force

$m a = W - 16 v m a = m g - 16 v 8 \frac{d v}{d t} = 8 (9.81) - 16 v 8 v' = 78.4 - 16 v$

Further solve the above expression,

role="math" localid="1663943376324" $v' = 9.8 - 2 v (Integrating factor e^{2 t}) v' + 2 v = 9.8 v . e^{2 t} = \int 9.8 e^{2 t} d t v . e^{2 t} = 4.9 e^{2 t} + C v = 4.9 e^{2 t} + C e^{- 2 t} v . e^{2 t} = 4.9 e^{2 t} + C$

At the value of $v = - 20 a n d t = 0, t h e n C = 24.9$ (c is constant and arrange negative sign)

$v = 4.9 + 24.9 e^{- 2 t}$

Find the equation of motion

$v = \frac{d x}{d t} \frac{d x}{d t} = 4.9 + 24.9 e^{- 2 t} x (t) = 4.9 t - 12.45 e^{- 2 t} + c$

Put the value of $t = 0, x = 100, t h e n C = 112.45$

$x (t) = 4.9 t - 12.45 e^{- 2 t} + 112.45$

Find the value of t

When the value of $x = 0$

$0 = 4.9 t - 12.45 e^{- 2 t} + 112.45$

By trial and error, the value of t $t = 22.9 \sec .$

Therefore, the value of t is $t = 22.9 \sec .$

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Short Answer

Step by step solution

Important hint.

Find the velocity

Find the equation of motion

Find the value of t

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