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)
A. 363 m/s
B. 402 m/s
C. 30.4 m/s
D. 164 m/s
Choose one answer
To calculate the ball's terminal velocity, we can use the equation:
v_t = √(2mg / (ρA*C))
where:
v_t = terminal velocity
m = mass of the ball
g = acceleration due to gravity (9.8 m/s^2)
ρ = air density
A = surface area of the ball
C = coefficient of drag
Plugging in the given values:
m = 0.15 kg
g = 9.8 m/s^2
ρ = 1.225 kg/m^3
A = 0.0026 m^2
C = 0.007
v_t = √(2*0.15*9.8 / (1.225*0.0026*0.007))
v_t = √(2.94 / 2.0653)
v_t = √1.4212
v_t ≈ 1.19 m/s
Therefore, the ball's terminal velocity is approximately 1.19 m/s. None of the given options match this value, so none of the given answers are correct.
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
F down = m g = 0.15 * 9.8 = 1.47 Newtons
F up = Cdrag * area * (1/2) rho v^2 = 0.007 * 0.0026 * 0.5 * 1.225 * v^2
so
1.47 = 0.000011147 v^2
v^2 = 1.47 *89706 = 131868
v = 363 m/s
which is A