Q.209. Mechanics, motion under gravity with air resistance.

Question 209.
A ball thrown vertically upwards with a velocity of 10 m/s returns to ground with a velocity of 9 m/s . What is the maximum height attained by the ball? [Take g = 9.8 m/s^2.]

Answer 209.
Since the velocity of the ball reduces on returning to the ground, it can be due to air resistance.

Let m = mass of the ball
and F = constant air resistance
=> deceleration in upward motion
= Net force / mass
= (mg + F) / m

If h = maximum height reached, where its velocity is zero,
0 = (10)^2 - 2 * [(mg + F) / m] * h
=> g + F/m = 50/h
=> F/m = 50/h - g ... (1)

Work done by air resistance = loss of kinetic energy
=> F * (2h) = (1/2) m [(10)^2 - (9)^2]
=> (F/m) * h = 19/4
=> F/m = 19/(4h) ... (2)

From (1) and (2),
50/h - g = 19/4h
=> g = 181/(4h)
=> h
= (181) / (4g)
= 4.62 m.

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