A ball of mass.0.4kg thrown upwards in air with a velocity 30m /s reaches the highest point in 2.5second. The air resistance encountered by
You need to know the radius of the ball in order to determine the answer to this question. Mass is irrelevant to the drag equations. It would also be helpful to know the density of the air because it may change due to temperature and humidity.
The basic equation of drag is:
F_D = \frac{1}{2} \rho v^2 C_D A
F_D is the drag force.
\rho is the density of the fluid (in this case, air).
v is the speed of the object relative to the fluid.
C_D is the drag coefficient. For a sphere, it is C_D = .47 . ( Drag coefficient - Wikipedia )
A is the cross sectional area.
Given the ball has radius r the area A = \pi r^2 , therefore, the drag is:
F_D = \frac{1}{2} \rho (30 m/s)^2 .47 (\pi r^2)
At sea level and at 15 °C, air has a density of approximately 1.225 \; kg/m^3 .
F_D = \frac{1}{2} 1.225 \; kg/m^3 (30 m/s)^2 .47 (\pi r^2)
Finally, we get,
F_D = 813.947 pascals * r^2
