A basketball star covers 3.00 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 3.00 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"). s

(b) Determine his horizontal velocity at the instant of takeoff. m/s

(c) Determine his vertical velocity at the instant of takeoff. m/s

(d) Determine his takeoff angle. ° above the horizontal

(e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, ymax = 2.60 m, and yf = 0.760 m.

Explanation / Answer

Given,

d = 3 m ; h(com) = 1.02 m ; Hm= 1.8 m; h' = 0.94 m

a)we know that

t = sqrt (2h/g)

t = sqrt [2x(1.9 - 1.02)/9.81] = 0.399 s

t' = sqrt[2(1.9 - 0.94)/9.81] = 0.442 s

T = t + t' = 0.399 + 0.442 = 0.841 s

Hence, T = 0.841 s

b)vx = D/t

Vx = 3/0.841 = 3.567 m/s

Hence, Vx = 3.567 m/s

c)using S = s0 + ut + 1/2 at^2

0.94 = 1.02 + Vy x 0.841 - 0.5 x 9.81 x 0.841^2

Vy = 4.03 m/s

d)theta = tan^-1(Vy/Vx)

theta = tan^-1(4.03/3.567) = 48.5 deg

Hence, theta = 48.5 deg

e) yi = 1.20 m, ymax = 2.60 m, and yf = 0.760 m.

t = sqrt [2x(2.6 - 1.2)/9.81] = 0.534 s

t' = sqrt[2(2.6 - 0.76)/9.81] = 0.612 s

T = t + t' = 0.534 + 0.612 = 1.15 s

Hence, T = 1.15 s

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