I've tried a bunch, have no idea. Got the first part, no idea on the second one.
A basketball, of mass m = 0.45 kg is dropped, from rest, from a stationary hellicopter at a high altitude. Use g = 9.81 N/kg.
(a) If air resistance is negligible, using energy considerations alone, how fast should the ball be going after it falls 100 meters? 1 m/s
(b) Of couse, we cannot ignore air resistance (i.e. drag). As a result, after the ball has fallen far enough, it reaches its terminal velocity, vt = �ãg/d, where d is the drag coefficient of the ball in air. If d = 0.007, how much energy has the ball lost due to air resistance after it falls 1300 meters?
(1/2) m v^2 = m g h
v^2 = 2gh
v = sqrt(2gh) = 44.3 m/s
Your drag coef makes no sense to me. Usually
Drag force = (1/2) rho v^2 Cd A
where rho = density of fluid (air here)
Cd = drag coef
A = cross section area of ball (pi r^2)
so at terminal speed
m g = (1/2) rho Vt^2 Cd A
Calculate Vt
calculate (1/2) m Vt^2 = Ke
now the Ke at the bottom would be
m g h = m g(1300) if there were no friction
so
work done by friction = 1300 m g - (1/2) m Vt^2
To solve part (a) of the problem, we can use the principle of conservation of energy. The initial potential energy of the ball is given by mgh, where m is the mass of the ball (0.45 kg), g is the acceleration due to gravity (9.81 N/kg), and h is the height (100 meters). The final kinetic energy of the ball is given by (1/2)mv^2, where v is the final velocity. Since no other energy losses are mentioned, we can assume that the potential energy is converted entirely into kinetic energy.
Setting up the equation:
mgh = (1/2)mv^2
Simplifying:
gh = (1/2)v^2
Solving for v:
v = sqrt(2gh)
Substituting the given values:
v = sqrt(2 * 9.81 * 100) ≈ 44.25 m/s
Therefore, the ball should be going approximately 44.25 m/s after falling 100 meters if air resistance is negligible.
To solve part (b) of the problem, we need to find the energy lost due to air resistance. The energy lost can be calculated by subtracting the final kinetic energy of the ball with air resistance from the initial kinetic energy.
The initial kinetic energy is given by (1/2)mv^2, where m is the mass of the ball (0.45 kg), and v is the initial velocity of the ball.
The final kinetic energy with air resistance can be calculated using the terminal velocity equation, given by vt = sqrt(gd), where g is the acceleration due to gravity (9.81 N/kg), and d is the drag coefficient of the ball in air (0.007). The final kinetic energy can be calculated as (1/2)mv_t^2, where m is the mass of the ball, and v_t is the terminal velocity.
The energy lost due to air resistance can be calculated as:
Energy lost = (1/2)mv^2 - (1/2)mv_t^2
Substituting the given values:
Energy lost = (1/2) * 0.45 * (44.25)^2 - (1/2) * 0.45 * (sqrt(9.81 * 0.007))^2
Using a calculator, we can evaluate this expression to find the value of the energy lost.
Therefore, the amount of energy lost due to air resistance after the ball falls 1300 meters can be calculated using the above formula.