Question

3. Derek throws a baseball upward from an initial height of 3 ft. The baseball hits the ground after 2 seconds.

Answer 1

Answer with Step-by-step explanation:

Since we have given that

Initial height = 3 feet

The baseball hits the ground after 2 seconds

t = 2 seconds

So, final velocity = v = 0

a. What was the initial velocity of the baseball?

Initial velocity is given by

$v_0=\dfrac{3}{2}=1.5\ ft/sec$

b. What is the function that models the height, ℎ (in feet), of the baseball over a period of time tt (in seconds)?

As we know that the function of height would be

$h(t)=-16t^2+v_0t+h_0$

Here, $h_0$ is the initial height.

So, our function becomes,

$h(t)=-16t^2+1.5t+3$

c. At what time did the baseball reach its maximum height? What was the maximum height of the baseball?

Maximum height at time t:

$t=\dfrac{-b}{2a}=\dfrac{-1.5}{2\times -16}=\dfrac{1.5}{32}\approx 0.05\ seconds$

Maximum height would be

$h(0.05)=-16(0.05)^2+1.5\times 0.05+3=3.035\ ft$