Questions: Question 3 1 pts A ball thrown vertically upward at time t=0 (seconds) has height y(t)= 96t-16t^2 (ft) at time t. Find the maximum height that the ball attains. Enter a single number for your answer. Height (in feet):

Question 3
1 pts

A ball thrown vertically upward at time t=0 (seconds) has height y(t)= 96t-16t^2 (ft) at time t. Find the maximum height that the ball attains.

Enter a single number for your answer.
Height (in feet):

Transcript text: Question 3 1 pts A ball thrown vertically upward at time $t=0$ (seconds) has height $y(t)=$ $96 t-16 t^{2}(\mathrm{ft})$ at time $t$. Find the maximum height that the ball attains. Enter a single number for your answer. Height (in feet): $\square$

Solution

Solution Steps

Step 1: Determine the time at which the maximum height is reached

To find the maximum height, we first need to determine the time $t$ at which it is reached. This is given by the formula $t = \frac{a}{2b}$, where $a = 96$ and $b = 16$. Substituting the given values, we get $t = \frac{96}{2 \times 16} = 3$ seconds.

Step 2: Calculate the maximum height

Having found the time, we can now calculate the maximum height $h_{max}$ using the formula $h_{max} = a(\frac{a}{2b}) - b(\frac{a}{2b})^2$. Substituting $a = 96$ and $b = 16$, we get $h_{max} = 96(\frac{96}{2 \times 16}) - 16(\frac{96}{2 \times 16})^2 = 144$ feet.

Final Answer:

The maximum height reached by the ball is ^144 feet^.

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