A basketball star covers 2.95 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.95 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 1.90 m above the floor end is at elevation 0.940 m when he touches down again. (a) Determine his time of flight (his 'hang time"). 8.70 (b) Determine his vertical velocity at the instant of takeoff. 4.15 m/s (c) Determine his vertical velocity at the instant of takeoff. 4.15 m/s (d) Determine his takeoff angle. 46 (e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations y_I = 1.20 M, Y_MAX = 2.40 m, and y_f = 0.770 m. s

Explanation / Answer

For horizontal displacement :

s = ut + 0.5*at^2

So, 2.95 = ux*t + 0 ------- (1)

For vertical displacement :

s = ut + 0.5*at^2

So, -(1.02 - 0.94) = uy*t + 0.5*(-9.8)*t^2 -------- (2)

Now, maximum height reached : h = uy^2/2g = (1.9 - 1.02)

So, uy = 4.15 m/s

So, from equation 2: t = 0.866 s <-----answer

b)

horizontal velocity at takeoff = ux

= 2.95/0.866 = 3.4 m/s <-----answer

c)

vertical velocity, uy = 4.15 m/s <---- solved above

d)

takeoff angle = atan(4.15/3.4) = 50.7 deg

e)

For hang time: we use the same equations :

h = uy^2/2g = (2.4 - 1.2)

So, uy = 4.85 m/s

So, -(1.2 - 0.77) = 4.85*t + 0.5*(-9.8)*t^2

So, t = 1.07 s <----- answer

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