basketball star covers 2.85 m horizontally in a jump to dunk the ball. His motio

Question

basketball star covers 2.85 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 1.06 m when he leaves the floor. It reaches a maximum height of 1.90 m above the floor and is at elevation 0.870 m when he touches down again. (Take the upward and forward directions to be positive.)
(a) Determine his time of flight (his "hang time").


(b) Determine his horizontal velocity at the instant of takeoff. (Indicate the direction with the sign of your answer.)


(c) Determine his vertical velocity at the instant of takeoff. (Indicate the direction with the sign of your answer.)


(d) Determine his takeoff angle.

Explanation / Answer

We know the final vertical velocity at any maximum height will be equal to zero: vfy^2 = (voy)^2 + 2ay 0 = (voy)^2 + 2(-9.8)(1.9-1.06) voy^2 = 16.464 voy = 4.06 m/s We can find the find "hang time" using the above velocity: y = (voy)(t) + (1/2)(a)(t^2) .870 - 1.06 = 4.06t + (1/2)(-9.8)(t^2) 4.9t^2 - 4.06t - .19 = 0 using quadratic formula: t = 0.873 seconds horizontal range: x = (vox)(t) 2.85 = (vox)(.873) vox = 3.26 m/s take off angle: angle = arctan(voy/vox) angle = arctan(4.06/3.26) angle = 51.23 degrees Answers: hang time = t = 0.873 seconds horizontal velocity at instant of takeoff = vox = 3.26 m/s vertical velocity at instant of takeoff = voy = 4.06 m/s take off angle = angle = 51.23 degrees bol

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