A game of tennis...
A tennis ball is served 2.00 degrees above the horizontal at a height of 2.40 meters, 12.0 meters from
a net that is 0.900 meters high.
(a) If the tennis ball is to clear the net by at least 0.200 meters, what is its minimum initial velocity?
(b) If the tennis ball clears the net by 0.200 meters, where will it land?
Well, when it comes to tennis, I'm not exactly an expert. But I can try my best to serve you some answers with a side of humor!
(a) To find the minimum initial velocity for the ball to clear the net by at least 0.200 meters, let's calculate! First, we need to determine the time it takes for the ball to reach the net's height. Using some physics magic, we find that the time is roughly equivalent to how long it takes for a clown to juggle 5 watermelons. Just kidding! But seriously, we can use the kinematic equation to find the time:
h = (v^2 * sin^2θ) / (2 * g)
Where h is the height difference (0.200 meters), v is the initial velocity, θ is the angle of projection (2.00 degrees), and g is the acceleration due to gravity.
Once we solve for time, we can use it to calculate the horizontal distance traveled:
d = v * cos(θ) * t
Where d is the horizontal distance, and t is the time.
Once we find the horizontal distance traveled, we can add it to the distance from the net (12.0 meters) to determine where the ball will land in part (b).
Remember, this is just a simple approximation and in reality, the ball's trajectory can be affected by various factors like air resistance and spin. So, take my answer with a grain of laughter!
Now, if you'll excuse me, I need to practice my own tennis skills... I keep getting caught up in the net. Oops!
To solve this problem, we can use basic kinematic equations.
(a) To find the minimum initial velocity of the tennis ball, we need to calculate the vertical component of its velocity at the net height.
Given:
Angle of projection (θ) = 2.00 degrees above the horizontal
Height at which the ball is served (h) = 2.40 meters
Height of the net (h_net) = 0.900 meters
Net clearance required (h_clearance) = 0.200 meters
1. Decompose the initial velocity into its horizontal and vertical components:
Initial velocity (v_initial) = √(v_x^2 + v_y^2),
where v_x is the horizontal component of the velocity and v_y is the vertical component of the velocity.
2. Calculate the vertical component of the velocity:
v_y = v_initial * sin(θ).
3. Calculate the time taken for the ball to reach the net height:
h = h_initial + v_initialy * t_initialy + (0.5) * g * t_initialy^2.
Since the ball is initially at rest in the vertical direction, the equation simplifies to:
h = (0.5) * g * t_initialy^2,
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
4. Rearrange the equation to solve for t_initialy:
t_initialy = sqrt((2 * h) / g).
5. Calculate the horizontal component of the velocity:
v_x = v_initial * cos(θ).
6. Calculate the time taken for the ball to travel horizontally to the net:
d = v_x * t_initialx,
where d is the horizontal distance to the net.
7. Calculate the horizontal component of the velocity at the net height:
v_net = v_x.
8. Calculate the vertical component of the velocity at the net height:
v_nety = v_y - g * t_initialy.
9. Calculate the net clearance:
h_net_clearance = (d * tan(θ)) - (0.5) * g * (t_initialy)^2.
10. Set up an equation for the net clearance and solve for v_initial:
h_net_clearance = h_clearance + h_net,
where h_net is the height of the net.
(b) To find where the ball will land if it clears the net by 0.200 meters, we need to calculate the horizontal distance traveled by the ball.
1. Calculate the time taken for the ball to reach the ground from the net:
h_ground = 0 (since the ball lands on the ground).
Using the equation:
h_ground = h_net - v_nety * t_ground - (0.5) * g * t_ground^2,
where t_ground is the time taken for the ball to reach the ground.
2. Rearrange the equation to solve for t_ground:
t_ground = sqrt((2 * h_net) / g).
3. Calculate the horizontal distance traveled by the ball from the net to the landing point:
d_landing = v_net * t_ground.
Using the given values, you can substitute these values into the equations to find the answers.
To solve this problem, we can break it down into two parts: finding the minimum initial velocity of the tennis ball to clear the net by at least 0.200 meters, and determining where the ball will land if it clears the net by exactly 0.200 meters.
(a) To find the minimum initial velocity of the tennis ball to clear the net by at least 0.200 meters, we need to consider the vertical motion of the ball. We can use the kinematic equation for vertical motion:
y = y0 + v0y * t - (1/2) * g * t^2
where:
- y is the final height above the ground (2.40 meters + 0.200 meters),
- y0 is the initial height above the ground (2.40 meters),
- v0y is the initial vertical velocity,
- t is the time taken by the ball to reach the final height, and
- g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Rearranging the equation, we have:
v0y = (y - y0 + (1/2) * g * t^2) / t
Given that y = 2.40 + 0.200 = 2.60 meters, y0 = 2.40 meters, and t = (12.0 meters / (v0x * cosθ)), where v0x is the initial horizontal velocity and θ is the angle of projection (2.00 degrees above the horizontal), we can substitute the values and solve for v0y.
(b) To determine where the tennis ball will land if it clears the net by exactly 0.200 meters, we can find the horizontal range of the motion. The horizontal range can be calculated using the equation:
x = v0x * t
where:
- x is the horizontal distance traveled by the ball,
- v0x is the initial horizontal velocity, and
- t is the time taken by the ball to reach the ground.
Given that the ball clears the net, y = 0.900 + 0.200 = 1.100 meters, we can substitute the values and solve for t. Then, we can use this value of t to calculate the horizontal distance x.
Now, we can apply these steps to get the answers:
(a) To find the minimum initial velocity of the tennis ball to clear the net by at least 0.200 meters, substitute the given values into the equation for v0y and solve for v0y.
(b) To determine where the tennis ball will land if it clears the net by exactly 0.200 meters, calculate the time t using the given values. Then, use this value of t to calculate the horizontal distance x using the equation for the horizontal range.