A basketball star covers 2.40 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.40 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 eight of 18 m a ove the floor, and is at elevation au m when he touches down again. (a) Determine his time of flight (his "hang time") Your response differs from the correct answer by more than 10%. Double check your calculations. (b) Determine his horizontal velocity at the instant of takeoff. 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. 3.9 m/s (d) Determine his takeoff angle Your response differs from the correct answer by more than 10%. Double check your calculations.° 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 = 2.50 m, and 0.690 m. lih.i-1 hmudieri ;, differs from the correct answer by more than 100%.

Explanation / Answer

PROJECTILE

along vertical
______________

initial velocity = v0y

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 1.02

from equation of motion

at maximum height

y = 1.8 m

vy = 0

vy^2 - voy^2 = 2*ay*(y-yo)

0 - voy^2 = -2*9.8*(1.8-1.02)

voy = 3.9 m/s

after touching down

y = 0.91

y - yo = voy*T + (1/2)*ay*Tt^2

0.91 - 1.02 = 3.9*T - (1/2)*9.8*T^2

time of flight T = 0.823 s <<<------------ANSWER

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(b)

along horizontal
________________

initial velocity = v0x

acceleration ax = 0

initial position = xo = 0

final position = x = 2.4

displacement = x - xo

from equation of motion

x - x0 = vox*T+ 0.5*ax*T^2

2.4 - 0 = vox*0.823

vox = 2.9 m/s

===============

(c)

vo = sqrt(vox^2+voy^2)

vo = 4.86 m/s

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(d)

angle = tan^-1(voy/vox) = 53.4 degrees

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(e)

PROJECTILE

along vertical
______________

initial velocity = v0y

acceleration ay = -g = -9.8 m/s^2

initial position yi = 1.2

from equation of motion

at maximum height

ymax = 2.5 m

vy = 0

vy^2 - voy^2 = 2*ay*(ymax - yi)

0 - voy^2 = -2*9.8*(2.5-1.2)

voy = 5.05 m/s

after touching down

yf = 0.69 m

yf - yi = voy*T + (1/2)*ay*Tt^2

0.69 - 1.2 = 5.05*T - (1/2)*9.8*T^2

time of flight T = 1.12 s <<<------------ANSWER

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