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.

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..