A basketball star covers 2.80 m horizontally in a jump to dunk the ball (see fig

Question

A basketball star covers 2.80 m horizontally in a jump to dunk the ball (see figure). 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 1.90 m above the floor and is at elevation 0.940 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

The vertical height he goes up is 1.90 -1.02 = 0.88m

Time to go up = t1 = ?[2h1/g] = ?[2*0.88/9.8]
0.42378s.

The vertical height he falls down is 1.90 -0.940 = 0.96m
Time to go up = t1 = ? [2h2/g] = ?[2*0.96 /9.8]
0.4426s

a)
Total time taken = 0.42378+0.4426 = 0.8664 s
b)
his horizontal velocity = horizontal distance / time
= 2.8 /0.8664 = 3.23173 m/s.
c)
The vertical velocity = u = ?[2gh] = ?[2*9.8*0.88]
=4.153m/s
d)
Take off angle is given by
tan ? = vertical component / horizontal
= 4.153/3.23173.
? = 52.1115?
======================================...
e)
this you can answer.

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