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.

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.