Question

A particle is thrown upward from ground, it experiences a constant air resistance 'f' and its weight 'w' find the speed with which it react at ground if velocity of projection is u

Answer 1

Answer:

Explanation:

In First Case, When the Ball is thrown upwards.

Initial Velocity = u

Final velocity(v) = 0

[∵ The ball will stop before coming down to the surface]

Height at which the ball reaches before coming down = S m

Time taken by the Particles to reach the height (or Time of Ascent) = t s.

Acceleration = -g

[∵ The ball is thrown against the gravity] 

From the Question, it experiences the air Resistance of 2 m/s².

∴ Acceleration of the Particles = -g - 2

 = - (g + 2) m/s².

Now, Using the Equation of the motion,

 v = u + at

∴ 0 + u + [-(g + 2)]t

∴ u = t(g + 2)

⇒ t = u/(g + 2)

Hence, time taken by the Particle to reach the certain height (or Time of Ascent) is u/(g + 2) seconds.

Now, Again using the Equation of the Motion,

v² - u² = 2aS

⇒ (0)² - (u)² = 2[-(g + 2)S]

⇒ u² = 2S(g + 2)

⇒ S = u²/(g + 2)

Now, In Second Case, (When the ball comes down),

Initial Velocity(u) = 0

Final Velocity = v

Time taken by the Particles to falls (or Time of Descent) = T s.

Height or  Distance covered(S) = u²/(g + 2) 

Acceleration of the Particle = g m/s²

∴ The Particles experienced the Air Resistance, 

 ∴ Acceleration = (g - 2) m/s².

∵ S = ut + 1/2 × aT²

∴ u²/(g + 2) = 0 × T + 1/2 × (g - 2) × T²

⇒ T = 

Hence, the Time taken by the Particles to falls down (or Time of Descent) is

Now, A/c to the Question,We have to calculate the ratio of the time of ascent to the time of descent.

∴ Time of Ascent/Time of Descent = t/T

   = u/(g + 2) ÷ u/√(g² - 4)

   = 

   = 

   = 

   = 

   = 

   = √96/12

   = 4√6/12

   = √6/3

∴ t : T = √6 : 3

Hence, the ratio of the time of the Ascent to the time of descent is √6 : 3