A basketball star covers 2.70 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.70 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 0.50 m when he touches down again. (a) Determine his time of flight (his ?hang time?). .8154 s (b) Determine his horizontal velocity at the instant of takeoff. 3.910 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (c) Determine his vertical velocity at the instant of takeoff. m/s (d) Determine his takeoff angle. degree 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 = 2.65 m, and yf = 0.710 m. __________ s

Explanation / Answer

a) voy = sqrt(2*g*(ymax-yi))

= sqrt(2*9.8*(1.8-1.02))

= 3.9 m/s

t_up = voy/g

= 3.9/9.8

= 0.399 s

(ymax-yf) = 0.5*g*t_down^2

d_down = sqrt(2*(ymax-yf)/g)

= sqrt(2*(1.8-0.95)/9.8)

= 0.416 s

T = t_up + t_down

= 0.8155 s

b) v0x = x/t

= 2.7/0.8154

= 3.31 m/

c) h = voy^2/(2*g)

voy = sqrt(2*g*h)

= sqrt(2*9.8*(1.8-1.02))

= 3.9 m/s

d) theta = tan^-1(voy/vox)

= tan^-1(3.9/3.31)

= 49.75 degrees

e) voy = sqrt(2*g*(ymax-yi))

= sqrt(2*9.8*(2.65-1.2))

= 5.33 m/s

t_up = voy/g

= 5.33/9.8

= 0.544 s

(ymax-yf) = 0.5*g*t_down^2

d_down = sqrt(2*(ymax-yf)/g)

= sqrt(2*(2.65-0.71)/9.8)

= 0.396 s

T = t_up + t_down

= 0.94 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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