To calculate the ball's terminal velocity, we can use the formula:
v_terminal = √(2 * m * g / (ρ * A * C_d))
Where:
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
C_d = coefficient of drag = 0.007
Plugging in the values, we get:
v_terminal = √(2 * 0.15 kg * 9.8 m/s^2 / (1.225 kg/m^3 * 0.0026 m^2 * 0.007))
v_terminal = √(2 * 0.147 kg*m/s^2 / 0.00002157 kg/m * 0.007)
v_terminal = √(13623.855 m^2/s^2 / 0.0001512 m^2/s)
v_terminal = √(90258730.74 / 151.2)
v_terminal = √(596439.42)
v_terminal ≈ 772.1 m/s
Since the options provided are not close to this value, it seems there may be an error in the calculation or options provided.
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) Responses
A. 363 m/s
B. 402 m/s
C. 30.4 m/s
D. 164 m/s
1 answer