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.

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach