Answer to Question #165545 in Mechanical Engineering for Nachikethas K

Question

An object with mass

m

= 0.6 kg is dropped from the top of a building and falls 21 m down in the ground

1.1 Calculate the speed of the object as it hits the ground and calculate

the time for the fall.

Neglect

in this question

the air resistance.

In a more accurate calculation we need to take air resistance into account.

In a wind tunnel the air resistance of the object has been measured giving

rise to an acceleration of:

air = 0.03v^2

where v is the speed of the object.

1.2 Now, taking air resistance into account, find the speed of the object as it

hits the ground. Any integrals in your calculation should be calculated

by hand!

1.3 Calculate the time of the fall when air resistance is taken into account.

Answer 1

Q165545

An object with mass

m = 0.6 kg is dropped from the top of a building and falls 21 m down in the ground

1.1 Calculate the speed of the object as it hits the ground and calculate the time for the fall.

Neglect in this question the air resistance.

Solution:

We will use the formula

$$v^2 = u^2 + 2 as$$

for finding the speed when the object just hits the ground.

Here 'v' is the velocity when the object hits the ground,

u is the initial velocity which in this case is equal to 0 m/s.

'a' is the acceleration due to gravity which is equal to 9.8 m/s².

's' is the distance traveled by the object.

plug u = 0m/s, a = 9.8 m/s² and s = 21 m in the formula we have

$v^2 = u^2 + 2 as = 0^2 + 2 * 9.8m/s^2 * 21m = 0 + 411.6 m^2/s^2 = 411.6 m^2/s^2$

$v^2 = 411.6 m^2/s^2$

Taking square root on both the side we have

$v = \sqrt{411.6m^2/s^2} = 20.29 m/s$

In 2 significant figures, the answer is 20. m/s

Hence the speed of the particle just when it hits the ground is 20.m/s

Next, we have to find the time

We will use the formula, v = u + a * t, for finding the time required by the particle to reach the ground.

Plug v = 20.29 m/s, u = 0 m/s and a = 9.8 m/s² in this formula

20.29 m/s = 0 m/ s + 9.8 m/s² * t

20.29 m/s = 9.8 m/s² * t

divide both the side by 9.8 m/s², we have

$\frac{20.29 m/s }{9.8m/s^2} = \frac{9.8 m/s2 * t }{9.8m/s^2}$

$t = \frac{20.29 m/s }{9.8m/s^2} = 2.070 seconds.$

which in 2 significant figures is 2.1 seconds.

Hence the object will take 2.1 seconds to reach the ground.

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In a more accurate calculation, we need to take air resistance into account.

In a wind tunnel, the air resistance of the object has been measured giving

rise to an acceleration of:

air = 0.03v^2

where v is the speed of the object.

1.2 Now, taking air resistance into account, find the speed of the object as it

hits the ground. Any integrals in your calculation should be calculated

by hand!

1.3 Calculate the time of the fall when air resistance is taken into account.

Solution:

The acceleration of air due to air resistance of object = 0.03 v ².

This acceleration is going to act opposite to the gravitational acceleration.

Hence the acceleration acting on the object will be = 9.8 m/s² - 0.03 v² ;

Substituting a = 9.8 m/s² - 0.03 v² in the formula v² = u² + 2 as, we have

v² = u² + 2 * ( 9.8 m/s² - 0.03 v² ) * s

Substitute u = 0m/s, and s = 21 m in this formula we have

v² = 0m/s ² + 2 * ( 9.8 m/s² - 0.03 v² ) * 21m

v² = 0 + 411.6 m²/s² - 1.26 v²

add 1.26 v² on both the side we have

v² + 1.26 v² = 0m/s ² + 411.6 m²/s²

2.26 v² = 411.6 m²/s²

divide both the side by 2.26 we have

$v^2 = \frac{411.6 m^2/s^2 }{2.26} = 182.124 m^2s^2$

Taking square root on both the side we have

$v = \sqrt{182.124 } = 13.50 m/s$

Hence when we consider the air resistance into consideration the object's velocity will be 13.50 m/s

when it just hits the ground.

Next we will substitute use the formula v = u + at , and find the time required for the object to reach

the group. here we will substitute a = 9.8m/s² + 0.03 * (13.50 m/s)² .

13.50 m/s = 0 m/s + ( 9.8m/s² - 0.03 * (13.50 m/s)² ) * t ;

13.50 m/s = 4.3325 m/s² * t

divide both the side by 4.3325 m/s² we have

$\frac{13.50 m/s}{4.3325 m/s^2 } = \frac{4.3325 m/s^2 * t }{4.3325 m/s^2 }$

$t = \frac{13.50 m/s}{4.3325 m/s^2 } = 3.116 seconds .$

which in 2 significant figures is 3.1 seconds.

Hence the particle will take 3.1 seconds to reach the ground if we take into account the resistance due to air.