A basketball star covers 2.85 m horizontally in a jump to dunk the ball (see fig

Question

A basketball star covers 2.85 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled as that of a particle at a point called his center of mass (which we shall define in Chapter 9). 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 and is at elevation 0.870 m when he touches down again.

(a) Determine his time of flight (his "hang time").
____ s

(b) Determine his horizontal velocity at takeoff.
____ m/s

(c) Determine his vertical velocity at takeoff.
____ m/s

(d) Determine his takeoff angle.
____ °

(e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, ymax = 2.55 m, and yf = 0.790 m.
____ s

Explanation / Answer

let initial vertical speed is v m/s

maximum height=1.9-1.02=0.88 m

at this maximum height, his vertical speed is 0.

then using final speed^2-initial speed^2=2*g*distance

where g=-9.8 m/s^2 (as it is opposed to the direction of motion)

we get,

0^2-v^2=-2*9.8*0.88

v=4.1531 m/s

time taken to reach maximum height=(final speed-initial speed)/acceleration

=(0-4.1531)/(-9.8)=0.42378 seconds

now, on the return path:

he will start will vertical speed of 0 m/s

acceleration=9.8 m/s^2 (positive because now it is acting along the direction of motion)

distance to cover=1.9-0.87=1.03 m

then using distance=initial speed*time+0.5*acceleration*time^2

we get time=0.45848 seconds

so net flight time=0.42378+0.45848=0.88226 seconds

part b:

horizontal velocity remains constant through out

hence horizontal velocity=horizontal distance covered/time taken

=2.85/0.88226=3.2303 m/s

part c:

vertical velocity=4.1531 m/s

part D:
take off angle=arctan(vertical speed/horizontal speed)=52.124 degrees

e)completing the same procedure as part a, we get hang time of the deer=1.1242 seconds

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