A tennis ball of mass 58g is thrown 5m into the air before served. Calculate the velocity of the tennis ball just before it is hit with the tennis racket.
We have to know how high the racket is when it hits the ball.
Maybe 3 meters?
if so the ball falls 2 meters from 5 meters to 3 meters
2 = 4.9 t^2
t = .639 seconds
v = g t = 9.8*.639 = 6.26 m/s
does not matter if on the way up or on the way down, same magnitude due to symmetry
Damon .. Pls view my question.. I need your opinion
V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6*5,
V = 9.90 m/s.
Note: It is assumed that the ball was hit at the same ht. it was released.
To calculate the velocity of the tennis ball just before it is hit with the tennis racket, we can use the principles of projectile motion and the conservation of energy.
Step 1: Convert the mass of the tennis ball from grams to kilograms:
1 g = 0.001 kg
58 g = 0.058 kg
Step 2: Determine the gravitational potential energy (GPE) of the ball at its maximum height:
Since the ball is thrown 5m into the air, at its maximum height its potential energy is fully converted from kinetic energy to gravitational potential energy.
GPE = m * g * h
where m is the mass (in kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (in meters).
GPE = 0.058 kg * 9.8 m/s^2 * 5 m
GPE = 2.843 J
Step 3: Use the conservation of energy to find the initial kinetic energy (KE) of the ball just before it is hit by the racket. The initial KE is equal to the GPE at maximum height.
KE = GPE
KE = 2.843 J
Step 4: Calculate the velocity using the formula for kinetic energy:
KE = 0.5 * m * v^2
where m is the mass (in kg) and v is the velocity (in m/s).
2.843 J = 0.5 * 0.058 kg * v^2
Step 5: Solve for v:
v^2 = (2.843 J) / (0.5 * 0.058 kg)
v^2 = 2.843 J / 0.029 kg
v^2 ≈ 98.034 m^2/s^2
Taking the square root of both sides:
v ≈ √(98.034 m^2/s^2)
v ≈ 9.901 m/s
Therefore, the velocity of the tennis ball just before it is hit with the tennis racket is approximately 9.901 m/s.