Question

An object having a mass of 5 kg is thrown vertically upward from a 15 m tower with an initial velocity of 7 m/sec. Air resistance acting on the object is equal to 6 10v.(a) Set up the differential equation required to solve the problem. Carefully define any variables you are using (with units) and state all appropriate initial conditions (with units) in terms of those variables.(b) When does the object reach its maximum height?

Answer 1

Step-by-step explanation:

The given data is as follows.

mass (m) = 5 kg

Height of tower = 15 m

u = 7 m/s

air resistance = 610 v

(a) Now, differential equation for the given mass which is thrown vertically upwards is as follows.

`m (d^(2)x)/(dt^(2))` = F

-bv = Fr

Here, mg is downwards due to the force of gravity.

`(md^(2)x)/(dt^(2)) = bv - mg`

`(md^(2)x)/(dt^(2)) + b (dx)/(dt) + mg` = 0

Hence, the differential equation required to solve the problem is as follows.

`(md^(2)x)/(dt^(2)) + b (dx)/(dt) + mg` = 0

(b) When final velocity of the object is equal to zero then the object will reach towards its maximum height and it will start to fall downwards.

F =
`(md^(2)x)/(dt^(2))`

= 0

Therefore, the object reach its maximum height at v = 0.