Question

a particle is moving along a projectile path at an initial height of 160 feet with an initial speed of 144 feet per second. this can be represented by the function $h(t) = -16t^2 + 144t + 160$. what is the maximum height of the particle?

• 128 feet
• 224 feet
• 272 feet
• 464 feet

Explanation:

Step1: Identify vertex of quadratic

The function $H(t) = -16t^2 + 144t + 160$ is a downward-opening quadratic, so its maximum is at the vertex. For $at^2+bt+c$, the $t$-coordinate of the vertex is $t=-\frac{b}{2a}$.

Step2: Calculate t-value of vertex

Substitute $a=-16$, $b=144$:
$$t = -\frac{144}{2(-16)} = \frac{144}{32} = 4.5$$

Step3: Compute max height at t=4.5

Substitute $t=4.5$ into $H(t)$:
$$H(4.5) = -16(4.5)^2 + 144(4.5) + 160$$
$$= -16(20.25) + 648 + 160$$
$$= -324 + 648 + 160$$

Answer:

484 feet