#1) (hrw8c2p64) A basketball player, standing near the basket to grab a rebound,

Question

#1) (hrw8c2p64) A basketball player, standing near the basket to grab a rebound, jumps 75.6 cm vertically.

(A) How much time does the player spend in the top 12.8 cm of his jump?
(B) How much time does the player spend in the bottom 12.8 cm of the jump?

#2) Given two vectors a = 3.0i -3.5j and b =6.6i+8.7j

(A) what is the magnitude (in m) of the vector a?
(B)Find the direction (in ° = deg) of the vector a.
(C)Find the magnitude of the vector b.
(D)Find the direction (in °) of the vector b.
(E)Find the magnitude of the vector a+b.
(F)Find the direction of the vector a+b.
(G)Find the magnitude of the vector b-a.
(H)Find the direction of the vector b-a
(J)Find the magnitude of the vector a-b.
(K)Find the direction of the vector a-b

Explanation / Answer

(A) The vertical velocity at the peak height is:

Vp = 0 at Hax = 0.756 m
using equn of motion
(Vp)² – (Vi)² = 2 • g • (Hax)
(0)² – (Vi)² = 2 • (-9.8) • (0.756)
Vi = 3.849 m/sec

The time to the peak of the jump is: Vp = (g • t) + Vi
0 = (-9.8) t + 3.849
t = 0.392 sec = Tp

The time to a height of 12.8 cm from the top of the jump is the time to a
height of:

Hax – 12.8 = 75.6 – 12.8 = 62.8 cm = 0.628 meters

D = (½) • g • t² + (Vi) • t
0.628 = (½) • (-9.8) • t² + (3.849) • t
0 = (-4.9) t² + (3.849) t – 0.628

Solutions: t = 0.23 sec and t = 0.55 sec

There are two solutions because there are two time values when the player is at
a height of 62.8 cm ... going up and going down.

So the total time the player is above a height of 62.8cm is conveniently the difference
between the two solution values:

0.55 – 0.23 = 0.32 sec = Top ... time in top portion of jump


(B) The total time in the air can be found using the same equation where the
final height (D) is zero:

D = (½) • g • t² + (Vi) • t
0 = (½) • (-9.8) • t² + (3.849) • t
0 = (-4.9) t² + (3.849) t
0 = t • [(-4.9)t + 3.849 ]

Solutions: t = 0 and (-4.9) t + 3.849 = 0
t = 0.7855 sec

so (t = 0) is at the start of the jump

and (t = 0.7855 sec = T total) is the total jump time from start to finish.

The time spent in the bottom 12.8 cm of the jump is:

To = Toa – Top
To = 0.7855 – 0.32
To = 0.4655 sec

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