how many seconds will pass before the ball hits the ground? h(t-80t^2+340t-260(in feet)

Question

I am looking for a start on this, thanks.

Answer 1

h(t)=-80t²+340t-260.

The ball reaches the ground when h(t)=0, so -80t²+340t-260=0.

The equation can be reduced by dividing by -20: 4t²-17t+13=0=(4t-13)(t-1).

So t=1 or t=13/4. That means t=1 second or 3¼ seconds. The answer is likely to be 3¼ seconds since t=1 second may be an initial condition when the ball was actually kicked or thrown. The number of seconds between the two events is 3¼-1=2¼ seconds.