052 A baseball mass 0145 kg is dropped from rest from a hot

05.2 A baseball (mass 0.145 kg) is dropped from rest from a hot air balloon. The magnitude of the drag force is Fdrag =(7.8 x 10-4 N-s 2 /m2 )v 2 , where v is the speed of the baseball.

(a) What is the terminal speed of the baseball?

(b) What happens is the baseball is thrown straight down with initial speed 50 m/s?

I am completely stuck on this question. I may be overthinking it.

Solution

a)

Fdrag = 7.8*10^-4*v^2

For terminal speed,

fdrag force must be equal to weight of the baseball

So, Fdrag = mg

So, 7.8*10^-4*v^2 = 0.145*9.8

So, v = 42.7 m/s <--------answer

b)

As the initial speed is greater than the terminal speed, the drag force will be greater than the force of gravity and this there is net deceleration of the ball.

So, its velocity will decrease upto 42.7 and then it remains with the same speed.