A ball of mass m is dropped from a height h above the ground, as shown in Figure 8.6. (a) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground.
Question : A ball of mass m is dropped from a height h above the ground...
- Mass of the ball: m
- Initial height of the ball: h
- Height of the ball above the ground: y
Step 1:
Identify the forces acting on the ball. In this case, the only force acting on the ball is the gravitational force.
Step 2:
Apply the principle of conservation of mechanical energy. Since the ball is in free fall, the total mechanical energy remains constant. Initially, the ball has potential energy but no kinetic energy. As the ball falls, its potential energy decreases while its kinetic energy increases, but the total mechanical energy remains the same.
Step 3:
Write the equation for conservation of mechanical energy. The initial potential energy of the system is equal to the final kinetic energy plus the final potential energy. This can be represented as: K_i + U_i = K_f + U_f, where K_i is the initial kinetic energy (which is 0), U_i is the initial potential energy, K_f is the final kinetic energy, and U_f is the final potential energy.
Step 4:
Substitute the given values into the equation. The initial potential energy is mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the initial altitude. The final kinetic energy is (1/2)mv_f^2, where v_f is the final velocity of the ball, and the final potential energy is mgy, where y is the distance above the ground.
Step 5:
Solve the equation for the final velocity. Rearrange the equation to solve for v_f: v_f^2 = 2g(h-y), and take the square root to find the final velocity.
Step 6:
Determine the sign of the final velocity. The speed is always positive, but if we were asked to find the ball's velocity, we would use the negative value of the square root to indicate the downward motion.
Step 7:
Repeat the process for the case where the ball already has an initial speed at the initial altitude. In this case, the initial energy includes the kinetic energy of the ball, and the equation for conservation of mechanical energy becomes (1/2)mv_i^2 + mgh = (1/2)mv_f^2 + mgy. Rearrange the equation and solve for v_f to find the speed of the ball at a given altitude y. This result is valid even if the initial velocity is at an angle to the horizontal.
Final Answer
Because the ball is in free fall, the only force act- ing on it is the gravitational force. Therefore, we apply the principle of conservation of mechanical energy to the ball – Earth system. Initially, the system has potential energy but no kinetic energy. As the ball falls, the total mechanical energy remains constant and equal to the initial potential en- ergy of the system. At the instant the ball is released, its kinetic energy is K_{i}=0 and the potential energy of the system is U_{i}=m g h . When the ball is at a distance y above the ground, its kinetic energy is K_{f}=\frac{1}{2} m v_{f}^{2} and the potential energy relative to the ground is U_{f}=m g y. Applying Equation 8.10, we obtain
K_{i}+U_{i} =K_{f}+U_{f} 0+m g h =\frac{1}{2} m v_{f}^{2}+m g y v_{f}^{2} =2 g(h-y) v_{f}=\sqrt{2 g(h-y)}The speed is always pasitive. If we had been asked to find the ball’s velocity, we would use the negative value of the square root as the y component to in dicate the downward motion.
(b) Determine the speed of the ball at y if at the instant of release it already has an initial speed v_{i} at the initial altitude h .
In this case, the initial energy includes kinetic energy equal to \frac{1}{2} m v_{i}^{2}, and Equation 8.10 gives \frac{1}{2} m v_{2}^{2}+m g h=\frac{1}{2} m v_{f}^{2}+m g y
v_{f}^{2} =v_{i}^{2}+2 g(h-y) v_{f} =\sqrt{v_{i}^{2}+2 g(h-y)}This result is consistent with the expression v_{y f}{ }^{2}=v_{y i}^{2}-2 g\left(y-y_{i}\right) from kinematics, where y_{i}=h . Furthermore, this result is valid even if the initial velocity is at an angle to the horizontal (the projectile situation) for two reasons: (1) energy is a scalar, and the kinetic energy depends only on the magnitude of the velocity; and (2) the change in the gravitational potential energy depends only on the change in position in the vertical direction.