Answer :
Answer:
Explanation:
Given
initial velocity of ball at ground [tex]u=80 m/s[/tex]
Distance traveled in t=3 sec is given by
[tex]h=ut+\frac{1}{2}at^2[/tex]
h=distance traveled
u=initial velocity
a=acceleration
t=time
[tex]h=80\times 3-\frac{1}{2}(-9.8)(3^2)[/tex]
[tex]h=195.9\ m[/tex]
Maximum Height is obtained when final velocity
using [tex]v^2-u^2=2 ah[/tex]
where v=final velocity
[tex]0-(80)^2=2\times (-9.8)\times h_{max}[/tex]
[tex]h_{max}=\frac{6400}{2\times 9.8}[/tex]
[tex]h_{max}=326.53\ m[/tex]
time taken to reach maximum height
using [tex]v=u+at[/tex]
[tex]0=80+(-9.8)\cdot t[/tex]
[tex]t=\frac{80}{9.8}=8.16 s[/tex]
therefore total time will be twice of time required to reach at maximum height
time taken to reach point D is [tex]2\times 8.16=16.32\ s[/tex]
