tball star covers 3.10 m horizontally in a jump to dunk the ball. His motion thr

Question

tball star covers 3.10 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 a) Determine his time of flight (his 'hang time"). b) Determine his horizontal velocity at the instant of takeoff mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of .80 m above the floor and is at elevation 0.910 m when he touches down agan. mys c) Determine his vertical velocity at the instant of takeoff. (d) Determine his takeoff angle. e) For con panson, de termine the hang tin e of a whtetail deer making a u p with center-of mass elevations y,1.20 m, m/s above the horizontal y ax 235 m, and n-o 700 m.

Explanation / Answer

(a) Here, when the basket ball star is at his maximum height, his vertical velocity is 0 m/s.

So, let us use the following equation to determine his initial vertical velocity.

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -9.8 m/s^2
d = 1.80 – 1.02 = 0.78 m
0 = vi^2 + 2 * -9.8 * 0.78

=> vi^2 = 15.288

=> vi = 15.288 = 3.91 m/s.

Now, let us use this number in following equation to determine the time to reach the maximum height.

vf = vi – a * t
0 = 3.91 – 9.8 * t

=> 9.8 * t = 3.91

=> t = 3.91 / 9.8 = 0.40 s.
Again -
1.80 – 0.91 = 0.89 m

So, let us use the following equation to determine the time for him to fall this distance.

d = vi * t + ½ * a * t^2, vi = 0, a = 9.8
0.89 = ½ * 9.8 * t^2
t = (0.89/4.9) = 0.43 seconds

Therefore, the time of flight = 0.40 + 0.43 = 0.83 second.

(b) Now during the time 0.83 s, he moved a horizontal distance of 3.10 meters. Let’s use the following equation to determine the horizontal component of his initial velocity.

d = v * t
3.10 = v * 0.83
v = 3.10 ÷ 0.83 = 3.73 m/s

(c) From above as we calculated -

Vertical velocity at the instant of takke off = 3.91 m/s

(d) Since we know the initial vertical and horizontal velocities, we can use the following equation to determine the angle.

Tan = Vertical ÷ Horizontal
Tan = 3.91 ÷ 3.73 = 1.05

So,   = 46.3 deg.

(e) For the white tail deer -

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -9.8 m/s^2
d = 2.35 - 1.20 = 1.15 m

so -
0 = vi^2 + 2 * -9.8 * 1.15
vi^2 = 22.54
vi = 22.54 = 4.75 m/s

again -

vf = vi – a * t
0 = 4.75 – 9.8 * t
9.8 * t = 4.75
t = 4.75 ÷ 9.8 = 0.48 second
again -
2.35 - 0.70 = 1.65 m
Let’s use the following equation to determine the time for him to fall this distance.

d = vi * t + ½ * a * t^2, vi = 0, a = 9.8
1.65 = ½ * 9.8 * t^2
t = (1.65/4.9) = 0.58 seconds
So, the total time of flight for the white tail deer = 0.48 + 0.58 = 1.06 second.

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