You serve a tennis ball from a height of 2.1 m above the ground. The ball leaves your racket with a speed of 19.2 m/s at an angle of 5.3° above the horizontal. The horizontal distance from the court's baseline to the net is 11.83 m, and the net is 1.07 m high. Neglect spin imparted on the ball as well as air resistance effects. Does the ball clear the net (= positive answer)? If yes, by how much? If not, by how much did it miss? In that case the answer will be negative. Answer in units of m
Vo = 19.2m/s[5.3o]
Xo = 19.2*cos5.3 = 19.11 m/s.
Yo = 19.2*sin5.3 = 1.77 m/s.
Tr=-Yo/g = -1.77/-9.8 = 0.181 s. = Rise
time.
h=ho + -Yo^2/2g = 2.1 + -(1.77^2/-19.6 = 2.1 + 0.160 m. = 2.26 m. = Max. ht. above gnd.
d = 0.5g*t^2 = 2.26 - 1.07 = 1.19 m.
4.9t^2 = 1.19
t^2 = 0.243
Tf = 0.493 s. = Fall time.
Dx = Xo*(Tr+Tf) = 19.11 * (0.181+0.493)=
12.88 m.
So the ball clears the net by:
12.88 - 11.83 = 1.05 m
To determine if the ball clears the net or not, we need to calculate the horizontal distance traveled by the ball and compare it to the distance to the net. If the horizontal distance is greater than the distance to the net, the ball clears the net.
To calculate the horizontal distance traveled by the ball, we need to decompose the initial velocity into its horizontal and vertical components. The horizontal component will remain constant throughout the ball's trajectory, while the vertical component will be affected by gravity.
The horizontal component of the initial velocity can be found using the formula:
Vx = V * cos(theta)
where V is the initial speed of the ball (19.2 m/s) and theta is the angle above the horizontal (5.3°).
Vx = 19.2 m/s * cos(5.3°)
Vx = 19.2 m/s * 0.99575 ≈ 19.128 m/s
Now, we can use the horizontal component of the initial velocity (Vx) to calculate the time it takes for the ball to reach the net. The time can be found using the formula:
t = d / Vx
where d is the distance to the net (11.83 m).
t = 11.83 m / 19.128 m/s ≈ 0.6176 s
Next, we can use the time (t) to calculate the vertical distance traveled by the ball using the formula:
Vy = V * sin(theta)
where V is the initial speed of the ball (19.2 m/s) and theta is the angle above the horizontal (5.3°).
Vy = 19.2 m/s * sin(5.3°)
Vy = 19.2 m/s * 0.08407 ≈ 1.613 m/s
Now, we can use the vertical component of the initial velocity (Vy) and the time (t) to calculate the vertical distance traveled by the ball using the formula:
Sy = Vy * t + (1/2) * g * t^2
where g is the acceleration due to gravity (9.8 m/s^2).
Sy = 1.613 m/s * 0.6176 s + (1/2) * 9.8 m/s^2 * (0.6176 s)^2
Sy = 0.9948 m + 1.8812 m ≈ 2.876 m
To determine if the ball clears the net, we need to compare the vertical distance traveled by the ball (Sy) to the height of the net (1.07 m).
If Sy > 1.07 m, then the ball clears the net by (Sy - 1.07 m). Otherwise, the ball misses the net by (1.07 m - Sy).
In this case, since Sy > 1.07 m (2.876 m > 1.07 m), the ball clears the net by 2.876 m - 1.07 m ≈ 1.806 m.
Therefore, the ball clears the net by approximately 1.806 m.