The robot has acceleration vector
[tex]\vec a=-g\,\vec\jmath[/tex]
where [tex]g=32\,\dfrac{\rm ft}{\mathrm s^2}[/tex] is the acceleration due to gravity.
Its initial velocity is
[tex]\vec v_0=\left(80\dfrac{\rm ft}{\rm s}\right)\cos80^\circ\,\vec\imath+\left(80\dfrac{\rm ft}{\rm s}\right)\sin80^\circ\,\vec\jmath[/tex]
so that its velocity at time [tex]t[/tex] is
[tex]\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm du[/tex]
[tex]\vec v=\left(\left(80\dfrac{\rm ft}{\rm s}\right)\cos80^\circ\,\vec\imath+\left(80\dfrac{\rm ft}{\rm s}\right)\sin80^\circ\,\vec\jmath\right)+\left(-gt\,\vec\jmath\right)[/tex]
[tex]\vec v=\left(80\dfrac{\rm ft}{\rm s}\right)\cos80^\circ\,\vec\imath+\left(\left(80\dfrac{\rm ft}{\rm s}\right)\sin80^\circ-gt\right)\,\vec\jmath[/tex]
If we take the robot's starting position to be the origin, so that [tex]\vec r_0=\vec 0[/tex], then its position vector at time [tex]t[/tex] is
[tex]\vec r=\vec0+\displaystyle\int_0^t\vec v\,\mathrm du[/tex]
[tex]\vec r=\left(80\dfrac{\rm ft}{\rm s}\right)\cos80^\circ\,t\,\vec\imath+\left(\left(80\dfrac{\rm ft}{\rm s}\right)\sin80^\circ\,t-\dfrac g2t^2\right)\,\vec\jmath[/tex]
The rocket is 10 feet horizontally away from its starting point when
[tex]\left(80\dfrac{\rm ft}{\rm s}\right)\cos80^\circ\,t=10\implies t=0.72\,\mathrm s[/tex]
At this point, its vertical position is
[tex]\left(80\dfrac{\rm ft}{\rm s}\right)\sin80^\circ(0.72\,\mathrm s)-\dfrac g2(0.72\,\mathrm s)^2=48\,\mathrm{ft}[/tex]
above the ground.
