Question

A ball is thrown with velocity v from ground in vertically upward direction if particle experiences resistance force mkv^2 where v is the speed of particle and mass of particle and K is a positive constant find maximum height reached​

Answer 1

Answer :

v²/2(g+kv²)

Solution :

• Given , a body is projected upwards with vertical initial velocity v
• Resistance force experienced = mkv²
• Also , gravity is exerted on the body , F due to gravity = mg
• Fnet downward = mg + mkv²
• Fnet = ma
• ma = m(g+kv²)
• a = g+kv² , downwards
• a = -(g+kv²)

Maximum height reached = ?

• Applying 3rd law of motion in a straight line for unifornly accelerated particles ,
• v²-u² = 2as
• At max point , v = 0 , s = h , a = g+kv² u = v
• 0²-v² = 2-(g+kv²)h
• -v² = -2(g+kv²)h
• h = v²/2(g+kv²)

Maximum height reached by the body is v²/2(g+kv²)

Answer 2

it has given that, initial velocity of ball = v ( upward). particle experience resistance force = mkv²

we have to find the maximum height reached by the particle.

solution : we know from Newton's 2nd law, F = ma

⇒-mg - mkv² = ma

⇒-g - kv² = a = v dv/dx

⇒-(g + kv²) = v dv/dx

⇒∫dx = -∫v/(g + kv²) dv

⇒x = -1/2k∫ 2kv/(g + kv²) dv

here if ( g + kv²) = f(x)

then, 2kv = f'(x)

we know, ∫f'(x)/f(x) dx = lnf(x) + C

⇒x = -1/2k [ln(g + kv²)]⁰_v [ final velocity becomes zero. because at highest point, velocity becomes zero ]

⇒x = 1/2k ln[(g + kv²)/g]

Therefore the maximum height reached by the ball is 1/2k ln[(g + kv²)/g]