sounds like a physics problem
basically we can simlpify it to this:
we toss something straight up with initial velocity 10ft/sec, assuming gravity kicks in, what is max height?
we can use the kinematic equation [tex]v_{f}^2=v_{0}^2+2ad[/tex]
[tex]v_f[/tex] is final velocity
[tex]v_0[/tex] is initial velocity
a is acceleration due to gravity
d=distance traveled
if we say he tosses it straight up then when it reaches max height, [tex] v_f=0[/tex]
and [tex]v_0=10 \frac{ft}{s}[/tex]
and we know that [tex]a=-32.174 \frac{ft}{s^2}[/tex]
so solving for d
[tex]v_{f}^2=v_{0}^2+2ad[/tex]
[tex]v_{f}^2-v_{0}^2=2ad[/tex]
[tex]\frac{v_{f}^2-v_{0}^2}{2a}=d[/tex]
plug the numbers in
[tex]\frac{(0)^2-(10 \frac{ft}{s})^2}{2(-32.174 \frac{ft}{s^2})}=d[/tex]
[tex]1.5540498539193137315845092310561 ft=d[/tex]
we add that to the initial 6ft
so total of 7.5540498539193137315845092310561 ft max height
about 7.6ft
