3. a toy rocket is launched from a platform that is 48 feet high. the rockets height above the ground is…

Explanation:

Step1: Identify the coefficients

The height - function is $h(t)=-16t^{2}+32t + 48$, where $a=-16$, $b = 32$, and $c = 48$.

Step2: Find the time of maximum height

The time $t$ at which the rocket reaches its maximum height is given by the formula $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 32$ into the formula: $t=-\frac{32}{2\times(-16)}=\frac{-32}{-32}=1$ second.

Step3: Find the maximum height

Substitute $t = 1$ into the height - function $h(t)=-16t^{2}+32t + 48$. $h(1)=-16\times(1)^{2}+32\times1 + 48=-16 + 32+48=64$ feet.

Step4: Find the time when the rocket reaches the ground

Set $h(t)=0$, so $-16t^{2}+32t + 48 = 0$. Divide the entire equation by -16: $t^{2}-2t - 3=0$. Factor the quadratic equation: $(t - 3)(t+1)=0$. Set each factor equal to zero: $t - 3=0$ or $t + 1=0$. Solving for $t$, we get $t = 3$ or $t=-1$. Since time cannot be negative, we discard $t=-1$.

Answer:

a) 64 feet b) 3 seconds