Question

anders throws a baseball straight up from a height of 3 feet above the ground. the initial velocity of the baseball is 25 feet per second. what is the height of the ball 1.5 seconds after it is thrown? round your answer to the nearest hundredth. (1 point)

Explanation:

Step1: Identify the height - time formula

The formula for the height $h(t)$ of an object in vertical - motion under the influence of gravity is $h(t)=-16t^{2}+v_{0}t + h_{0}$, where $v_{0}$ is the initial velocity, $h_{0}$ is the initial height, and $t$ is the time. Here, $h_{0}=3$ feet, $v_{0}=25$ feet per second, and $t = 1.5$ seconds.

Step2: Substitute the values into the formula

$h(1.5)=-16\times(1.5)^{2}+25\times1.5 + 3$.
First, calculate $-16\times(1.5)^{2}$:
$-16\times(1.5)^{2}=-16\times2.25=-36$.
Then, calculate $25\times1.5 = 37.5$.

Step3: Calculate the height

$h(1.5)=-36 + 37.5+3$.
$h(1.5)=-36+40.5$.
$h(1.5)=4.50$ feet.

Answer:

$4.50$ feet