A ball bounces upward from the ground with a speed of 14 m/s and hits a wall with a speed of 12 m/s. How high above the ground does the ball hit the wall? Ignore air resistance.
d = (Vf^2-Vo^2)/2g.
d = ((12)^2-(14)^2)/-19.6 = 2.65 m.
To determine the height above the ground where the ball hits the wall, we can use the principle of conservation of energy. Since we are assuming no air resistance, the mechanical energy of the ball is conserved throughout its motion.
Initially, the ball has kinetic energy due to its upward velocity. At the highest point in its trajectory, the ball has no velocity and therefore no kinetic energy. The only energy it possesses is potential energy.
The initial kinetic energy of the ball can be calculated using the formula:
KE_initial = (1/2) * m * v_initial^2
where m is the mass of the ball and v_initial is the initial velocity of 14 m/s.
When the ball hits the wall, its velocity has decreased to 12 m/s. Using the same formula, we can calculate the final kinetic energy:
KE_final = (1/2) * m * v_final^2
where v_final is the final velocity of 12 m/s.
Since mechanical energy is conserved, the initial kinetic energy is equal to the final potential energy of the ball at its highest point. The potential energy can be calculated using the formula:
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height above the ground.
Equating the initial kinetic energy to the final potential energy, we get:
(1/2) * m * v_initial^2 = m * g * h
Simplifying and rearranging the equation:
h = (v_initial^2 - v_final^2) / (2 * g)
Now we can substitute the given values into the equation to find the height:
h = (14^2 - 12^2) / (2 * 9.8)
h = (196 - 144) / 19.6
h = 52 / 19.6
h ≈ 2.65 meters
Therefore, the ball hits the wall approximately 2.65 meters above the ground.