A basketball star covers 2.45 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.45 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 1.75 m above the floor and is at elevation 0.870 m when he touches down again.

1. Determine his time of flight (his "hang time"). (answer in seconds)

2. Determine his horizontal velocity at the instant of takeoff. (answer in m/s)

3. Determine his vertical velocity at the instant of takeoff. (answer in m/s)

4. Determine his takeoff angle. ° above the horizontal

5. 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 = 0.800 m. (andswer in s)

Explanation / Answer

t1 = [2(1.75 - 1.02)/g] = 0.386 sec
t2 = [2(1.75 - 0.87)/g] = 0.4238 sec

a) Th = t = 0.8098 sec

b) Vx = (2.45/0.8098) = 3.02544 m/s

c) Vy = g*t1 = 3.7828 m/s

d) = arctan[Vy/Vx] = 51.35° above horizontal

e)  t1 = [2(2.6 - 1.2)/g] = 0.5345 sec
t2 = [2(2.6 - 0.8)/g] = 0.6061 sec

Th = t = 1.1406 sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.