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Wed 8 Aug, 2018 01:08 pm
A ping pong ball, of mass 2.7 g and diameter 4.0 cm, is dropped from a 15 m high building.
a. Estimate the ball’s terminal velocity.
b. What would be the impact velocity, with the ground below, in the absence of air drag?
@James2580,
To get terminal velocity, you need the drag coefficient and the density of the air. You can get the latter if you make some assumptions about temperature and elevation, but you can't solve (a) without the drag coefficient.
For b,
1/2 a t^2 = h = 15m
9.8m/sec^2 t^2 = 30m
t = 1.75 sec
v = a t
v = 9.8m/sec^2 * 1.75 = 17.1 m/s
@engineer,
engineer wrote:
To get terminal velocity, you need the drag coefficient and the density of the air. You can get the latter if you make some assumptions about temperature and elevation, but you can't solve (a) without the drag coefficient.
For b,
1/2 a t^2 = h = 15m
9.8m/sec^2 t^2 = 30m
t = 1.75 sec
v = a t
v = 9.8m/sec^2 * 1.75 = 17.1 m/s
You can get in the ball park results for such a light and slow moving object by just using civil engineering formula for wind pressure on a flat surface as follow
wind pressure per square foot = 0.00256 x the square of the wind speed.
I can remember using that formula to predict my terminal velocity when free falling with max body area exposed and it came out within a few MPH.
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