2.

Question

A ball is thrown into the air with an upward velocity of 24 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 24t + 7. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

0.75 s; 34 ft

1.5 s; 7 ft

0.75 s; 13 ft

0.75 s; 16 ft

Answer 1

To solve this problem, we need to find the maximum height of the ball, which occurs at the vertex of the parabola described by the quadratic function $h(t)=-16t^2+24t+7$.

The vertex form of a quadratic function $a{x}^2+bx+c$ has its vertex at $t=-\frac{b}{2a}$.

Here, $a=-16$ and $b=24$. Thus,

$$t=-\frac{24}{2(-16)}=-\frac{24}{-32}=\frac{24}{32}=\frac{3}{4}=0.75\text{ seconds}$$

To find the maximum height, we substitute $t=0.75$ back into the original function:

$$h(0.75)=-16(0.75{)}^2+24(0.75)+7$$

First, we calculate $(0.75{)}^2$:

$$(0.75{)}^2=0.5625$$

Next, we substitute and compute:

$$h(0.75)=-16(0.5625)+24(0.75)+7$$
$$h(0.75)=-9+18+7$$
$$h(0.75)=16$$

So, the ball's maximum height is 16 feet.

Therefore, the ball reaches its maximum height of 16 feet at $0.75$ seconds.

The correct answer is:

$$\boxed{{\displaystyle 0.75\text{ s};16\text{ ft}}}$$