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A shell of mass 10 kg is shot upward with an initial velocity of 200 ​m/sec. The magnitude of the force on the shell due to air resistance is v/20.
(a) When will the shell reach its maximum height above the​ ground? What is the maximum​ height? Assume the acceleration due to gravity to be 9.81 m/s^2.

Respuesta :

Answer:

A. 16.49 s.

B. 1536 m.

Explanation:

A.

Equation:

dv/dt + v/40 = −g

Where v(0)=200.

Integrating,

v(t) = e^(−t/40) * [200+40*g] −

40g.

At the maximum height, the object's final velocity = 0

Therefore,

solving for t,

t = −40 * ln (40g)/(200+40g)

= 16.49 s

B.

Remember,

v = dh/dt

Integrating,

v = dh/dt

= e^(−t/40) * (200+40*g) - 40*g

h(t) = 40 * (200) + 402*g − 40*(g*t) − e^(−t/40) * (40*(200)+402*g)

= 1536 m.