ch02-p064 - 64 The height reached by the player is y = 0.76...

2
0
2 2(9.8 m/s )(0.76 m) 3.86 m/s . v gy = = =
This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y
1
=
0.76 m – 0.15 m = 0.61 m, his speed v
1
satisfies v v gy
0
2
1
2
1
2 − = , which yields
2 2 2
1 0 1
2 (3.86 m/s) 2(9.80 m/s )(0.61 m) 1.71 m/s . v v gy = − = − =
The time t
1
that the player spends ascending in the top Δ y
1
= 0.15 m of the jump can now
be found from Eq. 2-17:
( )
( )
1 1 1 1
2 0.15 m
1
0.175 s
2 1.71m/s 0
y v v t t Δ = + ¡ = =
+
which means that the total time spent in that top 15 cm (both ascending and descending)
is 2(0.175 s) = 0.35 s = 350 ms.
(b) The time t
2
when the player reaches a height of 0.15 m is found from Eq. 2-15: