Question

Air resistance on particle can be considered to be uniform and directed opposite to the direction of its velocity in the vertical direction when a particle is thrown vertically upwards the time of ascent (t1) and time of descent(t2) are in the ratio 1:2 if g is the acceleration due to gravity the magnitude of uniform retardation by the air is ?

Answer 1

Do you know how to solve differential equations? I ask because a natural way to solve this question is write down the relevant DE and solve it.

If not, what formulae do you have for this situation?

Answer 2

R u talking abt calculus then i know it!

Answer 3

Let \( y \) be the positive vertical direction. Then the equation that governs this situation in the absence of air resistance is
\[ my'' = -mg \]
where ' is the time derivative. So far so good, yes?

Answer 4

i.e., F = ma = the force on the object from gravity

Answer 5

talk to me

Answer 6

@JamesJ it is a kinematics so we dont have to use force!

Answer 7

I'm trying to lead you to the point where we know what the relationship is between the coefficient of drag and the time it takes.

Answer 8

oh..sorry plz go on

Answer 9

Do you understand the equation I just wrote down?

Answer 10

F = ma and the only force acting on the mass here is gravity. Hence
ma = -mg
i.e.,
my'' = -mg

Now, in your problem, the air resistance is of uniform magnitude and in the opposite direction to the direction of motion. Call that magnitude K.

When the mass is rising, the force acts in the opposite direction of motion and hence is -K, thus the equation is

my'' = -mg - K

When the mass is falling ....

Answer 11

...the force is in the opposite direction of motion and hence in the opposite direction to gravity,

my'' = -mg + K

Answer 12

So now, given that the acceleration in both cases is a constant, you can use your normal kinematic equations to solve.

Find the time it takes to reach the top
then find the time it takes to come from the top back to the bottom.

Answer 13

i.e., when the object is going up, it has acceleration
a = y'' = -g - K/m

when the object is going down, it has acceleration
a = y'' = -g + K/m

Now use your kinematic equation
\[ y = y_0 + v_0t + \frac{1}{2}at^2 \]

Answer 14

i am confused @siddhantsharan

Answer 15

I think @JamesJ explained it extremely well.

Answer 16

still i am confused sice he used calculus can u use kinematics to solve it

Answer 17

@siddhantsharan i am waitind for ur answer

Answer 18

Let the deacceleration due to air be "a".
When particle goes up.
Net acceleration = -(g + a) (downwards)
v= 0
Let it have some initial velocity 'u'.
Therefore,
+s(some distance it goes up) = ut1 - 1/2(g+a)t1^2------(1)
And v = u + at. => 0 = u - (g+a)t1 => u = (g+a)t
Similarly on coming downwards,
Distance is same. Only direction is opposite. So h = -s.
Velocity at top = initial velocity = 0
Net Acceleration = -( g-a)
-s = -1/2(g-a)t2^2
s = 1/2(g-a)t2^2.------(2)
Equating 1 And 2. And substituting u = (g+a)t1. You should get your answer.

While solving divide the equation by t2^2 and substitude value for t1^2 / t2^2 = 1/4

You'll get your answer.