The basketball's total air time is approximately 1.514 seconds, reaching a maximum height of about 4.237 meters, with a horizontal range of approximately 2.618 meters. The final velocity just before hitting the ground is approximately 7.416 m/s.
a) Total air time (T):
T = 2v_{0y} / g
First, calculate v_{0y}:
v_{0y} = v_0 * sin(theta)
v_{0y} = 8.2 m/s * sin(65 degrees)
v_{0y} ≈ 8.2 m/s * 0.9063
v_{0y} ≈ 7.43 m/s
Now, substitute v_{0y} into the formula for T:
T = 2 * 7.43 m/s / 9.8 m/s^2
T ≈ 14.86 / 9.8
T ≈ 1.514 seconds
So, the total air time is approximately 1.514 seconds.
b) Maximum height (H):
H = y_0 + (v_{0y})^2 / (2 * g)
Substitute the known values:
H = 1.94 m + (7.43 m/s)^2 / (2 * 9.8 m/s^2)
H = 1.94 + 55.2049 / 19.6
H ≈ 4.237 m
So, the maximum height is approximately 4.237 meters.
c) Range (R):
R = v_{0x} * T / 2
First, calculate v_{0x}:
v_{0x} = v_0 * cos(theta)
v_{0x} = 8.2 m/s * cos(65 degrees)
v_{0x} ≈ 8.2 m/s * 0.4226
v_{0x} ≈ 3.461 m/s
Now, substitute v_{0x} into the formula for R:
R = 3.461 m/s * 1.514 seconds / 2
R ≈ 5.236 / 2
R ≈ 2.618 meters
So, the range is approximately 2.618 meters.
d) Final velocity (v_f):
v_f = sqrt((v_{0x})^2 + (v_{0y} - gt)^2)
Substitute the known values at the time of impact (t = T):
v_f = sqrt((3.461 m/s)^2 + (7.43 m/s - 9.8 m/s^2 * 1.514 s)^2)
v_f ≈ sqrt(11.95321 + (-6.6367)^2)
v_f ≈ sqrt(11.95321 + 44.06369)
v_f ≈ sqrt(55.0169)
v_f ≈ 7.416 m/s
So, the final velocity is approximately 7.416 m/s.