A tennis player hits a ball 1.6 m above the ground. The ball leaves

the racket with a speed of 12 m/s at an angle 8.0 degrees above the horizontal. The net is a horizontal distance of 7.0 m away from the player and is 1.0 m high.
(a.) Does the ball clear the net? Explain how you determined this. You must do a
correct numerical calculation as part of your explanation.
(b.) If the ball clears the net, by what distance does it clear it? If not, by what distance
does it miss?

To determine if the ball clears the net, we need to calculate the horizontal distance it would travel.

(a.) The initial velocity of the ball can be split into its horizontal and vertical components. The horizontal velocity (Vx) remains constant throughout the motion, while the vertical velocity (Vy) changes due to gravity.

Given:
Initial velocity (V) = 12 m/s
Launch angle (θ) = 8.0° (above the horizontal)
Initial height (h) = 1.6 m
Distance to the net (d) = 7.0 m
Net height (H) = 1.0 m

First, let's calculate the time it takes for the ball to reach the net:
Using the vertical component of the initial velocity:
Vy = V * sin(θ)
Vy = 12 * sin(8°)
Vy ≈ 1.65 m/s

To calculate the time of flight (t), we can use the kinematic equation:
h = (1/2) * g * t^2
1.6 = (1/2) * 9.8 * t^2
t^2 ≈ 0.3265
t ≈ √(0.3265)
t ≈ 0.57 s

Now, let's calculate the horizontal distance covered by the ball:
Using the horizontal component of the initial velocity:
Vx = V * cos(θ)
Vx = 12 * cos(8°)
Vx ≈ 11.97 m/s

d = Vx * t
d = 11.97 * 0.57
d ≈ 6.82 m

Since the horizontal distance covered by the ball is less than the distance to the net (6.82 m < 7.0 m), the ball does not clear the net.

(b.) The ball misses the net by a distance equal to the difference between the horizontal distance covered by the ball and the distance to the net:
Distance missed = d - 7.0 m
Distance missed ≈ 6.82 m - 7.0 m
Distance missed ≈ -0.18 m (It misses the net by 0.18 m to the left of the net)

To determine whether the tennis ball clears the net, we need to calculate the trajectory of the ball and compare it to the net height.

(a.) To calculate the trajectory, we can split the initial velocity into horizontal and vertical components.

The vertical component of the velocity can be calculated as:
Vy = V * sin(theta)
where V is the initial speed of 12 m/s and theta is the angle of 8.0 degrees.

The horizontal component of the velocity can be calculated as:
Vx = V * cos(theta)

Considering the vertical motion, we can use the equation:
h = (Vy^2) / (2 * g)
where h is the height above the ground (1.6 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Solving for Vy, we get:
Vy = sqrt(2 * g * h)

Plugging in the values, we find:
Vy = sqrt(2 * 9.8 * 1.6) = 6.31 m/s

The time it takes for the ball to reach its maximum height is given by:
t = Vy / g

Plugging in the values, we find:
t = 6.31 / 9.8 = 0.645 seconds

Now, considering the horizontal motion, we can calculate the horizontal distance the ball travels in this time:
x = Vx * t
where x is the horizontal distance and Vx is the horizontal component of the velocity.

Plugging in the values, we find:
x = 12 * cos(8.0) * 0.645 = 7.76 m

Since the horizontal distance (7.76 m) is greater than the distance to the net (7.0 m), we can conclude that the ball will clear the net.

(b.) To calculate the distance by which the ball clears the net, we need to determine the vertical distance between the height of the net and the height of the ball's trajectory.

The vertical distance can be calculated as:
vertical_distance = (height of the ball's trajectory) - (height of the net)

Plugging in the values, we find:
vertical_distance = 1.6 - 1.0 = 0.6 m

Therefore, the ball clears the net by a distance of 0.6 meters.

a. Vo = 12m/s[8o]

Xo = 12*cos8 = 11.88 m/s.
Yo = 12*sin8 = 1.67 m/s.

Tr = -Yo/g = -1.67/-9.8 = 0.170 s. = Rise time.

h = ho + (Y^2-Yo^2)/2g
h = 1.67 + (0-(1.67^2))/-19.6 = 1.74 m.
Above gnd.

d = 0.5g*t^2 = 1.74-1.0
4.9*t^2 = 0.74
t^2 = 0.151
Tf = 0.389 s. = Fall time.

Dx = Xo * (Tr+Tf)=11.88 * (0.170+0.389)=
6.64 m = Hor. distance. So it falls short of the net.

b. Missed by: 7-6.64 = 0.36 m.