2points
an athlete jumps with a speed of 4 m/s at an angle of 23º. how long does the
athlete stay in the air?
o a. 0.32 s
o b. 0.925
o c. 0.55 s
o d. 0.75 s

A. 0.32s

Explanation:

Lets find the initial velocity of the vertical component

Cos23=u/4

4cos23=1.6m/s

then time, t =2u/g

2×1.6/9.8

=0.32seconds

A. 0.32 s

Explanation:

The other answer did the math wrong but came up with the same answer. Anyway, you're supposed to use sin.

4sin(23) gives you 1.56 or 1.6

-1.56/-9.8 (negative because gravity is working against the direction of of the athlete as he goes into the air) = 0.159...

0.159*2 = 0.318

Round: 0.32

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can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s