Question

Answer the following questions: A basketball star covers 3.10 m horizontally in a jump to dunk the ball. His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.80 m above the floor and is at elevation 0.940 m when he touches down again.

a. Determine his time of flight (his "hang time").

b. Determine his horizontal velocity at the instant of takeoff.

c. Determine his vertical velocity at the instant of takeoff.

d. Determine his takeoff angle.

e. For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, ymax = 2.35 m, and yf = 0.740 m.

Answer 1

(A)

yi = 1.02 m and y_max= 1.80 m

vf^2 - vi^2 = 2 a y

0^2 - v0y^2= 2(-9.8)(1.80 - 1.02)

v0y = 3.91 m/s

for time of flight, yf = 0.940 m

yf - yi = v0y t + ay t^2 / 2

0.940 - 1.02 = 3.91 t - 4.9 t^2

4.9 t^2 - 3.91 t - 0.08 =0

t = 0.818 sec

(B) v0x = 3.10 / 0.818 = 3.79 m/s

(C) v0y = 3.91 m/s

(D) theta = tan^-1(v0y / v0x) = 46 deg

(E) v0y = sqrt(2 x 9.8 x (2.35 - 1.20)) = 4.75 m/s

0.740 - 1.20 = 4.75 t - 4.9 t^2

4.9 t^2 - 4.75 t - 0.46 = 0

t = 1.06 sec