A ball falls from a tall bridge. The ball has a mass of 0.15 kg. The air density is 1.225 kg/m3
. The ball’s surface area is 0.0026 m2
. The ball’s coefficient of drag is 0.007. What is the ball’s terminal velocity?(1 point)
1 402 m/s
2 363 m/s
3 164 m/s
4 30.4 m/s
The equation for terminal velocity is given by:
Vt = sqrt((2mg)/(ρACd))
where:
Vt = terminal velocity
m = mass of the ball (0.15 kg)
g = acceleration due to gravity (9.8 m/s^2)
ρ = air density (1.225 kg/m^3)
A = surface area of the ball (0.0026 m^2)
Cd = coefficient of drag (0.007)
Plugging in the values into the equation:
Vt = sqrt((2 * 0.15 * 9.8) / (1.225 * 0.0026 * 0.007))
Vt = sqrt(0.294 / 0.0000019665)
Vt = sqrt(149521.61)
Vt ≈ 386.29 m/s
Therefore, the ball's terminal velocity is approximately 386.29 m/s.
The closest option is 2) 363 m/s.