A tennis player standing 11.5 m from the net hits the ball at 2.60° above the horizontal. To clear the net, the ball must rise at least 0.252 m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racquet?
To solve this problem, we can use the projectile motion equations. The horizontal and vertical motions are independent of each other, so we can analyze them separately.
First, let's consider the vertical motion of the ball. We know that the ball needs to rise at least 0.252 m to clear the net. We can use the formula for the vertical displacement of a projectile:
Δy = v₀y * t + (1/2) * g * t²
where:
Δy is the vertical displacement (0.252 m),
v₀y is the initial vertical velocity (unknown),
t is the time of flight (unknown),
g is the acceleration due to gravity (-9.8 m/s²).
Since we want to find the initial vertical velocity, we can rewrite the equation as:
v₀y = (Δy - (1/2) * g * t²) / t
Next, let's consider the horizontal motion of the ball. We can assume there is no air resistance, so the horizontal velocity remains constant throughout the flight. The horizontal displacement can be found using the formula:
Δx = v₀x * t
where:
Δx is the horizontal displacement (11.5 m),
v₀x is the initial horizontal velocity (unknown),
t is the time of flight (unknown).
Rearranging this equation, we can solve for the initial horizontal velocity:
v₀x = Δx / t
Now, we need to determine the time of flight, which is the same for both vertical and horizontal motion. We can use the equation for the vertical motion to find the time when the ball reaches its maximum height (apex). At the apex, the vertical velocity is zero.
v_y = v₀y + g * t
Since v_y = 0 at the apex, we can solve for the time of flight:
0 = v₀y + g * t
t = -v₀y / g
Now we can substitute this value of t into the equations for v₀y and v₀x to find the initial vertical and horizontal velocities, respectively.
v₀y = (Δy - (1/2) * g * (-v₀y / g)²) / (-v₀y / g)
v₀x = Δx / (-v₀y / g)
Simplifying these equations:
v₀y = (Δy - (1/2) * v₀y² / g) / (-1 / g)
v₀x = Δx * (g / v₀y)
Now we have two equations with one unknown (v₀y). We can solve them simultaneously to find the initial vertical velocity.
Let's substitute the known values into the equations and solve for v₀y:
Δy = 0.252 m
Δx = 11.5 m
g = -9.8 m/s²
v₀y = (0.252 - (1/2) * v₀y² / (-9.8)) / (-1 / (-9.8))
v₀x = 11.5 * (-9.8 / v₀y)
Now we can solve these equations simultaneously.
0.252 + (1/2) * v₀y² / 9.8 = 9.8 * v₀y
0.252 * 9.8 + (1/2) * v₀y² = 9.8² * v₀y
(1/2) * v₀y² - 9.8² * v₀y + 0.252 * 9.8 = 0
Solving this quadratic equation using the quadratic formula:
v₀y = (-(-9.8²) ± √((-9.8²)² - 4 * (1/2) * 0.252 * 9.8)) / (2 * (1/2))
v₀y = (9.8² ± √(9.8⁴ - 4 * 0.252 * 9.8)) / 1
v₀y = (9.8² ± √(9.8⁴ - 4 * 0.252 * 9.8)) / 1
Calculating this expression gives two possible solutions for v₀y: 53.7 m/s and 9.22 m/s. Since the ball is hit upwards, the initial vertical velocity should be positive, so v₀y = 53.7 m/s.
Now we can substitute this value of v₀y into the equation for v₀x to find the initial horizontal velocity:
v₀x = 11.5 * (-9.8 / 53.7)
v₀x = -2.102 m/s
Therefore, the ball was moving with an initial velocity of approximately -2.102 m/s horizontally and 53.7 m/s vertically when it left the racquet.
To find the speed of the ball when it left the racquet, we can use the principle of conservation of energy. The initial kinetic energy of the ball is converted into gravitational potential energy as it rises to its highest point.
First, let's calculate the vertical distance the ball traveled. Given that the ball cleared the net by 0.252 m and the net is 11.5 m away from the player, we can use the tangent function to find the vertical distance traveled.
We have:
tangent(2.60°) = (vertical distance traveled) / (horizontal distance traveled)
tan(2.60°) = vertical distance / 11.5 m
Rearranging the equation, we get:
vertical distance = tan(2.60°) * 11.5 m
Now, we know that at the highest point of the ball's trajectory, the vertical distance is maximum. So, the highest point is 0.252 m above the net. Therefore, we equate the vertical distance to 0.252 m:
tan(2.60°) * 11.5 m = 0.252 m
Next, we can solve for the initial velocity of the ball using the kinematic equation for vertical motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s when the ball reaches the highest point)
u = initial velocity (what we are solving for)
a = acceleration (gravitational acceleration, -9.8 m/s^2)
s = vertical distance (0.252 m)
Substituting the values, we get:
0^2 = u^2 + 2 * (-9.8 m/s^2) * 0.252 m
0 = u^2 - 4.992 m^2/s^2
Rearranging and solving for u, we have:
u^2 = 4.992 m^2/s^2
u = √(4.992 m^2/s^2)
Therefore, the initial vertical velocity of the ball is:
u = √(4.992 m^2/s^2)
To find the initial velocity of the ball, we need to use the horizontal component of the velocity. Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant throughout the ball's flight.
Since we know the ball cleared the net at the apex of its trajectory, we can consider that its horizontal velocity doesn't change. Therefore, the horizontal component of the initial velocity is the same as the horizontal component of the final velocity.
To find the horizontal component of the final velocity, we can use the equation:
v_x = u_x
Where:
v_x = final horizontal velocity (what we are solving for)
u_x = initial horizontal velocity (what we are solving for)
Now, we need to find the time it takes for the ball to reach the highest point of its trajectory. We can use the vertical component of the initial velocity to calculate the time:
v = u + at
Where:
v = final vertical velocity (0 m/s when the ball reaches the highest point)
u = initial vertical velocity (what we found earlier)
a = vertical acceleration (gravitational acceleration, -9.8 m/s^2)
t = time
Substituting the values, we get:
0 = √(4.992 m^2/s^2) + (-9.8 m/s^2) * t
Simplifying, we have:
√(4.992 m^2/s^2) = 9.8 m/s^2 * t
Squaring both sides of the equation:
4.992 m^2/s^2 = (9.8 m/s^2)^2 * t^2
Rearranging and solving for t, we have:
t^2 = 4.992 m^2/s^2 / (9.8 m/s^2)^2
t = √(4.992 m^2/s^2) / (9.8 m/s^2)
Now, we can find the horizontal distance traveled by the ball using the time calculated:
horizontal distance = horizontal velocity * time
Substituting the values, we have:
11.5 m = horizontal velocity * (√(4.992 m^2/s^2) / 9.8 m/s^2)
Rearranging and solving for the horizontal velocity, we get:
horizontal velocity = 11.5 m * (9.8 m/s^2) / √(4.992 m^2/s^2)
Finally, to find the initial velocity of the ball when it left the racquet, we can use the Pythagorean theorem:
initial velocity = √(horizontal velocity^2 + vertical velocity^2)
Substituting the values we calculated earlier, we get:
initial velocity = √(horizontal velocity^2 + (√(4.992 m^2/s^2))^2)
Simplifying, we have:
initial velocity = √(horizontal velocity^2 + 4.992 m^2/s^2)