Answer:8.1 m
Explanation:
Given
ball is launched from height of 3 m
initial velocity [tex]u=10 m/s[/tex]
considering the ball is thrown vertically upward
Using [tex]v^2-u^2=2 as[/tex]
where,
u=initial velocity
v=final Velocity
a=acceleration
s=distance
At maximum height final velocity will be zero
[tex]v^2-u^2=2 gs[/tex]
[tex]0-10^2=2(-9.8)\cdot h[/tex]
[tex]h=5.10 m[/tex]
Therefore maximum height w.r.t ground is [tex]3+5.10=8.10 m[/tex]
