IIT JEE , AIEEE Forum - Physics , Mechanics

sucess_seeker

In an imaginary atmosphere. the air exerts a small force F on any particle in the direction of the particle's motion. A particle of mass m is projected upward takes a time t₁and reaching the maximum height and t₂ in the return journey to the original point. then how is t₁>t₂.

i think both the time should be equal as time of ascent = time of descent

vasanth

let the vertical height travelled be H

a_upwards= g - F/m

u is initial velocity(of projection)

t₁ = mu/(mg -F)

when the particle starts its downward motion the atmosphere exerts F in the downward direction

a_downwards= F/m + g

the final velocity is v

t₂ = mv/(mg + F)

v is greater than u as there is an external force, so energy of the mass is not conserved.

v² = 2(g + F/m)H

u² = 2(g - F/m)H

v² = u² (mg + F)/(mg - F)

= u² {1 + 2F/(mg - F)}

v = u {1 + F/(mg - F)}---------binomial expansion

t₂ = mu(1 + F/(mg - F))/(mg + F)

= mu(mg)/(mg + F)

so this method does not give a determinable answer.

SECOND METHOD:

H = ut₁ + 1/2(g - F/m)t₁²

t₁² (g - F/m) + 2ut₁ - 2H = 0

t₁ = -2u +-

(4u² + 8H(g- F/m))/2(g- F/m)

= -u +

(u² + 2H a_upwards)/ a_upwards

t₁² = 2u² + 2H a_up - 2u

(u² + 2H a_upwards)/a²_up

t₁² = 2(u/a_up)²+ 2H/u_up - 2u

(u² + 2H a_upwards)/a²_up

when the particle starts its downward motion the atmosphere exerts F in the downward direction

H = 1/2(a_downwards)t₂²

t₂² = (2H)/a_downwards

Compare the two eqns and get the answer

I'm in a bit o' hurry......i'll complete the soln later......

phanindraramesh

You are correct by saying that t₁>t₂

because the air resistance exist the body experiences both air resistance and gravitational force in opposite to it's motion

but while coming down the force of gravitation is along the motion of the body

air resistance is opposite to the motion of body

Hence it wil take more time for reaching maximum height and less time to come down

So t₁>t₂

casio_ss

hey its simple one.....

look while going mg will be opposite to that of the motion. hence it will take more time to reach at Hmax.

while descending, mg will be in direction of the motion hence...it will take less time.....

that force F is same as atmospheric pressure but the only think is it is not opposing the motion...

do rate me!!!!!

sucess_seeker

thank u everybody

elessar_iitkgp

Let the initial speed be u and height reached be h.
For the upward journey,
0 = u + (F/m - g)T₁
T₁ = u / (g -F/m) ---------(1)

And

0 = u² +2(F/m - g)h
h = u² /2(g - F/m) ---(2)

For the downward journey,
-h = (1/2)[ -(F/m + g)]T_2²T₂ =

[(2u² /2(g - F/m)/)/(F/m + g)] (Substituting the value of h from (2))
T₂= u/

(g² - (F/m)²) -------(3)

Now,
(g + F/m)>(g - F/m)
Multiplying both sides by (g - F/m)
(g² - (F/m)²)>(g - F/m)²

(g² - (F/m)²)>(g - F/m)
u/

(g² - (F/m)²)<u / (g -F/m)
T₂<T₁

punterjack

t1>t2. net force during t1=mg-F. during t2=mg+F as u can see net force during t2>that during t1 as a result of which the same length can be covered in lesser time hence t1>t2 HP

omkarnet

your problem concist of air resitance therefore in jee we have to negelct air resitance

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