a ball is thrown up on the surface of a moon. its height above the lunar surface (in feet) after t seconds…

Explanation:

Step1: Identify the function type

The height - function $h(t)=490t-\frac{10}{4}t^{2}=490t - 2.5t^{2}$ is a quadratic function of the form $y = ax^{2}+bx + c$, where $a=-2.5$, $b = 490$, and $c = 0$.

Step2: Find the time for maximum height

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case, the time $t$) of the vertex is given by $t=-\frac{b}{2a}$. Substitute $a=-2.5$ and $b = 490$ into the formula: $t=-\frac{490}{2\times(-2.5)}=\frac{490}{5}=98$ seconds.

Step3: Find the maximum height

Substitute $t = 98$ into the height - function $h(t)=490t-2.5t^{2}$. $h(98)=490\times98-2.5\times98^{2}$ $h(98)=490\times98-2.5\times9604$ $h(98)=48020 - 24010$ $h(98)=24010$ feet.

Answer:

Answer for the time that the ball reaches its maximum height: 98 Answer for the maximal height attained by the ball: 24010