A basketball star covers 2.80 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.80 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 1.85 m above the floor, and is at elevation 0.900 m when he touches down again.
Determine
(a) his time of flight (his “hang time”),
(b) his horizontal and
(c) vertical velocity components at the instant of takeoff, and
(d) 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 = 2.50 m, yf = 0.700 m.

Explanation / Answer

? the equation of star’s behaviour in constant gravity is parabola in parametric form: x(t) = u*t, y(t) =-0.5g*t^2 +v*t +hi, where x(t) is horizontal component of his position, y(t) is his vertical coordinate, u is initial and constant horizontal component of his speed, v is initial vertical component of his speed, hi=1.02m, time t being the parameter; ? thus x(t) =2.8m =u*t, hence t=2.8/u; also y(t) = 0.91m = -0.5*g*t^2 +v*t +1.02; now substituting t: 0.5g(2.8/u)^2 –v*(2.8/u) –0.11 =0; ? besides he’s given vertex: 1.95m = -0.5g*t’^2 +v*t’ +hi, where t’ is time to reach vertex; his vertical component of speed being 0 = -g*t’ +v, hence t’ = v/g; therefore 1.95 =-0.5g*(v/g)^2 +v*(v/g) +1.02; 2*0.93*9.8 = v^2, hence v=4.2694 m/s; ? return to (?): 0.5g(2.8/u)^2 – 4.2694*(2.8/u) –0.11 =0; 0.5g*2.8^2 - 4.2694*2.8*u –0.11u^2 =0; u^2 +108.676u –349.2364 =0, u=3.1215 m/s; his take-off angle = atan(u/v) =atan(3.1215/4.2694) =36.17°; hang time =2.8/u = 0.897 s; ¦ comparison; return to (?): 2*(2.7-1.2)*9.8 =v^2; v=5.4222 m/s; return to (?): 0.76m= -0.5g*t^2 +v*t +1.2; t^2 –2*5.4222*t/9.8 –2*(1.2-0.76)/9.8 =0; t=1.1825s; also good!

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