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?

To solve this problem, we can use the equations of motion and the principles of projectile motion. Let's break down the problem step by step:

(a) To find the minimum initial velocity, we need to determine the speed at which the ball leaves the racquet. We can use the equation:

v^2 = u^2 + 2as

where:
v = final velocity (0 m/s since the ball reaches maximum height)
u = initial velocity (what we're trying to find)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
s = displacement (vertical distance, which is the difference in height)

Given the displacement s = 2.40 m - 0.900 m + 0.200 m = 1.700 m, and since the ball reaches its maximum height (v = 0 m/s), we can rearrange the equation to solve for u:

u^2 = v^2 - 2as
u^2 = 0^2 - 2(-9.8 m/s^2)(1.7 m)
u^2 = 33.32 m^2/s^2
u = √(33.32 m^2/s^2)
u ≈ 5.77 m/s

Therefore, the minimum initial velocity needed for the tennis ball to clear the net by at least 0.200 m is approximately 5.77 m/s.

(b) To find where the ball will land, we need to determine the horizontal distance it travels. We can use the equation:

d = ut + 0.5at^2

where:
d = horizontal distance (what we're trying to find)
u = initial velocity (5.77 m/s)
a = acceleration (0 m/s^2 since there is no horizontal acceleration)
t = time of flight (what we're trying to find)

Since the ball is served at an angle above the horizontal, we can decompose the initial velocity into horizontal and vertical components:

u_x = u * cosθ
u_y = u * sinθ

where:
θ = angle above the horizontal (2.00 degrees in this case)

Given u = 5.77 m/s and θ = 2.00 degrees, we can calculate the horizontal and vertical initial velocities:

u_x = 5.77 m/s * cos(2.00 degrees)
u_y = 5.77 m/s * sin(2.00 degrees)

Now we can determine the time of flight using the vertical motion equation:

y = u_y * t + 0.5 * a * t^2

where:
y = vertical displacement (height of the net, which is 0.900 m)

Plugging in the values:

0.900 m = (5.77 m/s * sin(2.00 degrees)) * t + 0.5 * (-9.8 m/s^2) * t^2

This is a quadratic equation in t, which can be solved to find the time of flight (t).

Once we have the time of flight, we can calculate the horizontal distance using:

d = u_x * t

Therefore, by calculating the time of flight and multiplying it by the horizontal component of the velocity, we can determine where the ball will land.