While dunking a ball a basketball player covers 2.80 m horizontally. In the foll

Question

While dunking a ball a basketball player covers 2.80 m horizontally. In the following problem model the athlete as a single point located at the player's center-of-mass. Just as the athlete leaves the ground their center of mass Ls located 1.02 m above the ground. In the course of the flight, the player's center of mass reaches a maximal height of 1.85 m, and as the player lands the center of mass Ls located 0.900 m above the ground. Determine the player's time of flight (i.e. hang time). Determine the vector components of the player's initial velocity as the jump begins. Determine the takeoff angle with respect to the ground. Compare this to the hang time of a whitetail deer making a jump with center-of-mass elevations y_o = 1.20 m, y_max = 2.50 m, and y_f = 0.700 m.

Explanation / Answer

a) at maximum height, vertical component of velocity becomes zero.

using , vf^2 - vi^2 = 2a y

0^2 - (v sin@)^2 = 2 (-9.8)(1.85 - 1.02)

vsin@ = 4.03 .......(i)

In vertical using, y = uy*t + ay*t^2 /2

(0.900 - 1.02) = (4.03)t + (-9.8 t^2 / 2)

4.9t^2 - 4.03t - 0.12 = 0

t = 0.85 sec ......Ans

b) in horizontal, he travels 2.80m

hence vx = Dx / t = 2.80 / 0.85 = 3.29 m/s

and vy = 4.03 as calculated in a.

hence v = 3.29i + 4.03j

c) tan@ = 4.03/ 3.29

@ = 50.77 deg

d) using , vf^2 - vi^2 = 2a y

0^2 - (v sin@)^2 = 2 (-9.8)(2.50 - 1.20)

vy = vsin@ = 5.05 .......(i)

In vertical using, y = uy*t + ay*t^2 /2

(0.7 - 1.20) = (5.05)t + (-9.8 t^2 / 2)

4.9t^2 - 5.05t - 0.5 = 0

t = 1.12 sec ......Ans

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