Answer :
Given data:
The expression for the height is h= -16t^2 +vt+c.
The expression for the velocity is,
[tex]V=-32t+v[/tex]At t=0, the velocity is 50ft/s.
[tex]v=\text{ 50 ft/s}[/tex]At t=0, h= 4ft
[tex]c=4[/tex]The expression for h is,
[tex]h=-16t^2+50t+4[/tex]Substitute 10 ft for h in the above expression.
[tex]\begin{gathered} 10=-16t^2+50t+4 \\ 16t^2-50t+6=0 \\ t=3\text{ s, 0.125 s} \end{gathered}[/tex]Here, two value of times indicates that one in the ascent and second one in the descent.
Thus, the value of time are 3 seconds and 0.125 s.