A basketball star covers 2.75 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.75 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 0.950 m when he touches down again.
(a) Determine his time of flight (his "hang time").
s

(b) Determine his horizontal velocity at the instant of takeoff.
m/s

(c) Determine his vertical velocity at the instant of takeoff.
m/s

(d) Determine his takeoff angle.

Explanation / Answer

t1 = ?[2(1.75 - 1.02)/9.8] = 0.385 sec
t2 = ?[2(1.75 - 0.95)/9.8] = 0.404 sec

a) Th = ?t = 0.789 sec

b) Vx = (2.75/0.789) = 3.485 m/s

c) Vy = g*t1 = 9.8*0.385 = 3.773 m/s

d) ? = arctan[3.773/3.485] = 47.27 degree above horizontal

e) 2*(2.3-1.2)*9.8 =v^2;

v=4.643 m/s;

0.65m= -0.5g*t^2 +v*t +1.2;
t^2

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