A tennis player serves a tennis ball it is moving horizontally when it leaves the racquet. When the ball travels a horizontal distance of 12 meters, it has dropped 55cm from its original height when it left the racquet. What is the initial speed of the ball? (neglect air resistance)
See previous post: Sun, 9-13-15, 4:17 PM.
To find the initial speed of the tennis ball, we can use the principle of conservation of energy. The potential energy that the ball loses as it drops is converted into kinetic energy.
First, let's determine the height from which the ball was served. We are given that the ball dropped 55cm from its original height when it left the racquet. This means the height it was served from is 55cm higher than where it landed.
Next, we can calculate the initial potential energy of the ball. The potential energy formula is given by:
Potential Energy = mass * gravity * height
In this case, we neglect air resistance, so we can assume the mass of the ball remains constant. The acceleration due to gravity is approximately 9.8 m/s^2.
Now, let's find the mass of the ball. Since we only have the information about the horizontal distance the ball travels, we can't directly determine the mass. However, we can make an assumption that the initial vertical velocity is zero, which means there is no initial momentum in the vertical direction. Therefore, we can assume that the ball is in freefall for 55cm and determine the mass of the ball using the formula:
Kinetic Energy = 1/2 * mass * velocity^2
When the ball reaches its maximum height (55cm), its potential energy is equal to the kinetic energy it had when it left the racquet. Rearranging the formula, we have:
Potential Energy = Kinetic Energy
mass * gravity * height = 1/2 * mass * velocity^2
Canceling mass on both sides and rearranging the equation, we get:
velocity^2 = 2 * gravity * height
Now we can plug in the values and calculate the initial speed:
velocity^2 = 2 * 9.8 m/s^2 * 0.55 m
velocity^2 = 9.62 m^2/s^2
Taking the square root of both sides, we find:
velocity = √(9.62 m^2/s^2) ≈ 3.10 m/s
Therefore, the initial speed of the ball is approximately 3.10 m/s.