An inflated spherical beach ball with a radius of 0.4000 m and average density of 10.65 kg/m3 is being held under water in a pool by Janelle. The density of the water in the pool is 1000.0 kg/m3. When Janelle releases the ball, it begins to rise to the surface. If the drag coefficient of the ball in the water is 0.520 and the constant upward force on the ball is 2630 N, what will be the terminal speed of the ball as it rises

3.71 m/s

Explanation:

Given data :

radius of the spherical ball = 0.400 m

Density of the ball material  = 10.65 Kg/m^3

Density of water = 1000 Kg/m^3

Drag coefficient C_d= 0.520

Constant upward force  F_up= 2630

Concept: The moment before it starts to rise Upward force must be equal to downward drag force

⇒ F_up = F_D

F_D =C_d× 0.5 ×ρ_w×V^2×A

where V= terminal velocity and A=  area πR^2

equating the forces we get

2630= 0.52× 0.5×1000×V^2×π(0.4)^2

V= 3.171 m/s

Therefore, the terminal velocity of the ball = 3.71 m/s

it would be kevlar ^^