Solution Given: h(x)=40−10x2h(x)=40-10x^2h(x)=40−10x2 (i) Heights at given times h(1)=40−10(1)2=30h(1.5)=40−10(1.5)2=40−10(2.25)=17.5h(1.7)=40−10(1.7)2=40−10(2.89)=11.1\begin{aligned} h(1)&=40-10(1)^2=30\\ h(1.5)&=40-10(1.5)^2=40-10(2.25)=17.5\\ h(1.7)&=40-10(1.7)^2=40-10(2.89)=11.1 \end{aligned}h(1)h(1.5)h(1.7)=40−10(1)2=30=40−10(1.5)2=40−10(2.25)=17.5=40−10(1.7)2=40−10(2.89)=11.1 h(1)=30 m,h(1.5)=17.5 m,h(1.7)=11.1 m\boxed{h(1)=30\text{ m},\quad h(1.5)=17.5\text{ m},\quad h(1.7)=11.1\text{ m}}h(1)=30 m,h(1.5)=17.5 m,h(1.7)=11.1 m (ii) When does the stone strike the ground? Set h(x)=0h(x)=0h(x)=0: 40−10x2=0⇒10x2=40⇒x2=4⇒x=240-10x^2=0\Rightarrow 10x^2=40\Rightarrow x^2=4\Rightarrow x=240−10x2=0⇒10x2=40⇒x2=4⇒x=2 (Time is non-negative.) x=2 seconds\boxed{x=2\text{ seconds}}x=2 seconds