Given the equation: -m g + k v^2 = m a
Find:
A) The time it takes for a golf ball dropped from an elevation of 1000 feet to hit the ground.
B) Velocity of the golf ball when it hits the ground.
C) What is the terminal velocity of the golf ball?
m=mass of golf ball
g=32.17 ft/s^2
k=(Cd)(rho)(ball frontal area)/2 = 4.60 x10^-6
v=velocity
a=acceleration
W=0.10 lb (golf ball weight), where W=mg, or m=W/g
Assume drag coefficient (Cd) and air density (rho) are constant for simplicity, therefore k is constant.
The term k v^2 is the drag force due to wind acting upward when the ball falls downward (wind drag force is proportional to the square of the velocity).
Hints:
a = dv/dt
v = dy/dt, or the integral of v(t)=y(t) where y is the elevation.

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