A player passes a 0.600-kg basketball down court for a fast break. The ball leaves the player's hands with a speed of 8.20m/s and slows down to 7.30m/s at its highest point.Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?
0.941m
.711m
bobpursley always be giving the wrong answer
53
1.06
To determine the height above the release point where the basketball reaches its maximum height, we need to use the principles of energy conservation.
First, let's determine the gravitational potential energy of the basketball at its highest point. The gravitational potential energy (PE) is given by the equation:
PE = m * g * h,
where m is the mass of the basketball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the release point.
Next, let's consider the kinetic energy of the basketball at its highest point. The kinetic energy (KE) is given by the equation:
KE = 0.5 * m * v^2,
where v is the velocity of the basketball at its highest point.
According to the principle of energy conservation, the total mechanical energy (the sum of potential and kinetic energy) remains constant. Therefore, the initial mechanical energy (E_i) is equal to the final mechanical energy (E_f) at the highest point:
E_i = E_f.
Initially, the basketball is released with a velocity of 8.20 m/s, so the initial kinetic energy is:
KE_i = 0.5 * m * v_i^2,
where v_i = 8.20 m/s.
At the highest point, the basketball's velocity is 7.30 m/s, so the final kinetic energy is:
KE_f = 0.5 * m * v_f^2,
where v_f = 7.30 m/s.
The final potential energy is:
PE_f = m * g * h.
Setting the initial mechanical energy equal to the final mechanical energy, we have:
KE_i + PE_i = KE_f + PE_f.
Substituting the equations for kinetic and potential energy, we get:
0.5 * m * v_i^2 + m * g * h = 0.5 * m * v_f^2 + m * g * h.
Simplifying, we can cancel out the mass (m):
0.5 * v_i^2 + g * h = 0.5 * v_f^2 + g * h.
Rearranging the equation, we solve for the height (h):
h = (0.5 * v_f^2 - 0.5 * v_i^2) / g.
Substituting the given values:
v_i = 8.20 m/s,
v_f = 7.30 m/s,
g = 9.8 m/s^2,
we can now calculate the height (h).
Ok, the change in KE goes to PE
1/2 m (8.2^2-7.3^2)=mgh
solve for h.