This question is incomplete the complete question is
A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?
Answer:
(a) Xs=0.459m
(b) t=0.984 s
(c) Vc=6.65 m/s
Explanation:
(a) To reach maximum distance
[tex]g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\ x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m[/tex]
(b) For Time
To find t we must find t1 and t2
as
t=t1+t2
For T1
[tex]t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s[/tex]
For T2
[tex]x_{l}=Vbt+(1/2)gt_{2}^{2}\\ as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s[/tex]
For Total Time
t=t1+t2
t=0.306+0.6789
t=0.984s
(c) To find Vc
Vc=Vb+gt2
Vc=(0)+(9.8)(0.6789)
Vc=6.65 m/s
