Question

Hello,

A soccer player kicks a ball straight up in the air at 18 m/s.
What is the balls total hang time? 3.68 s
What is the maximum height reached by the ball? 16.53 m
When is the ball 12 m above the ground?

Vavg = (v+vnaught)/2
Vavg= (18+0)/2
Vavg=9m/s
d=vavg*t
12m=9t
t=1.33s

21.06m=9m/s(t)
t=2.34s

Is that correct?

Thanks!

Answer 1

Hang tiime:
first, time to top.
vf=vi-gt
0=18-9.8t >>t= 18/9.8=1.84 sec
hang time, double that or 3.68 sec
max height:
vf^2=vi^2-2*9.8*h
h= 18^2/19.6=16.53 m

time to get to 12m.
d=vi*t-1/2 g t^2
12=18t-4.8 t^2
put that in quadratic form, use quadratic equation
4.8t^2-18t+12=0
t^2-3.67t+1.22=0
t= (3.6-+sqrt(3.6^2-4.88))/2=1.8+-1.42 =1.32 sec at 12 meter mark
check all that.

Answer 2

1. V = Vo + g*Tr = 0.
18 + (-9.8)Tr = 0,
Tr = 1.84 s. = Rise time.
Tf = Tr = 1.84 s. = Fall time.
Tr + Tf = 1.84 + 1.84 = 3.68 s. = Total time in air or hang time.

2. V^2 = Vo^2 + 2g*h = o.
18^2 + (-19.6)h = 0,
h = 16.53 m.