Answer :
notice, it hits the ground when d = 0, thus
[tex]\bf \stackrel{d}{0}=-16t^2+55\implies 16t^2=55\implies t^2=\cfrac{55}{16}\implies t=\sqrt{\cfrac{55}{16}}[/tex]
that's when.
as far as its graph, is more or less like the picture below, now, notice that if we put the quadratic in "vertex form", it'd look like
[tex]\bf d=-16t^2+55\implies d=-16(t-\stackrel{h}{0})^2+\stackrel{k}{55}\qquad vertex~(h,k)[/tex]
so its vertex is at (0, 55), so just pick one point to the left of the vertex, and one to the right of the vertex, and draw it away
[tex]\bf \stackrel{d}{0}=-16t^2+55\implies 16t^2=55\implies t^2=\cfrac{55}{16}\implies t=\sqrt{\cfrac{55}{16}}[/tex]
that's when.
as far as its graph, is more or less like the picture below, now, notice that if we put the quadratic in "vertex form", it'd look like
[tex]\bf d=-16t^2+55\implies d=-16(t-\stackrel{h}{0})^2+\stackrel{k}{55}\qquad vertex~(h,k)[/tex]
so its vertex is at (0, 55), so just pick one point to the left of the vertex, and one to the right of the vertex, and draw it away
