h=0 h=3+13t-5t^2

Answer 1

h=0= 3+13t-5t^2

graph the parabola and see where the curve intersects the x axis

it hits the ground when t= 2.813 unts of time

or use the quadratic formula t= -b/2a+/-(1/2a)sqr(b^2-4ac)

=t= 13/10 +/-.1sqr(169+60)

= 1.3 +/-(sqr229)/10 = 1.3+1.513= about 2.813, ignoring negative time

it starts at height of 3, then reaches max height of 11.45 at t=1.3,

then falls down reaching h=0 when t=2.813