Answer:
Step-by-step explanation:
Given
initial Velocity(u)=44 ft/s
ball initial height =4 ft
as launch angle is not given so considering ball is hit vertically upward
[tex]v^2-u^2=2as[/tex]
where u =initial velocity
v=final velocity
a=acceleration
s=distance traveled
[tex]0-44^2=2\times (-32.2)\cdot s[/tex]
[tex]s=\frac{44^2}{2\times 32.2}=30.06 ft[/tex]
time taken to reach 30.06 ft
[tex]v=u+at[/tex]
[tex]0=44-32.2\times t[/tex]
[tex]t=1.36 s[/tex]
total distance to reach ground=4+30.06 ft=34.06 ft
[tex]s=ut+\frac{at^2}{2}[/tex]
[tex]34.06=\frac{32.2\times t^2}{2}[/tex]
[tex]t^2=2.115[/tex]
t=1.45 s
