A ball is thrown so that its height is modeled by h(t) = -16t² + 40t, where h is in feet and t is in seconds. At what time, other than…

Set the height to zero: $-16t^2+40t=0$. Factor out $t$: $t(-16t+40)=0$, so $t=0$ or $-16t+40=0$. The second equation gives $t=\frac{40}{16}=2.5$ seconds, choice C. The ball is thrown at $t=0$ and lands at $t=2.5$. Check: $-16(6.25)+40(2.5)=-100+100=0$. Choice B ($2.0$) gives $-64+80=16$ feet, so the ball is still in the air. Choice A ($1.5$) gives $-36+60=24$ feet. Choice D ($4.0$) gives $-256+160=-96$, which is below the ground and not physically meaningful. The ball is in the air only between the two zeros of the height function.