QuestionApril 30, 2025

A projectile with an initial velocity of 48 feet persecond is What is the maximum height of the projectile? launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t)=-16t^2+ 48t+190 82 feet 190 feet 226 feet 250 feet

A projectile with an initial velocity of 48 feet persecond is What is the maximum height of the projectile? launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t)=-16t^2+ 48t+190 82 feet 190 feet 226 feet 250 feet
A projectile with an initial velocity of 48 feet persecond is What is the maximum height of the projectile?
launched from a building 190 feet tall. The path of the
projectile is modeled using the equation h(t)=-16t^2+
48t+190
82 feet
190 feet
226 feet
250 feet

Solution
4.5(267 votes)

Answer

226 feet Explanation 1. Find the time at maximum height The maximum height occurs at the vertex of the parabola. Use the formula for the time at the vertex: t = -\frac{b}{2a}, where a = -16 and b = 48. t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5 seconds. 2. Calculate the maximum height Substitute t = 1.5 into the equation h(t) = -16t^2 + 48t + 190: h(1.5) = -16(1.5)^2 + 48(1.5) + 190 h(1.5) = -16(2.25) + 72 + 190 h(1.5) = -36 + 72 + 190 = 226 feet.

Explanation

1. Find the time at maximum height<br /> The maximum height occurs at the vertex of the parabola. Use the formula for the time at the vertex: $t = -\frac{b}{2a}$, where $a = -16$ and $b = 48$. <br />$t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5$ seconds.<br /><br />2. Calculate the maximum height<br /> Substitute $t = 1.5$ into the equation $h(t) = -16t^2 + 48t + 190$:<br />$h(1.5) = -16(1.5)^2 + 48(1.5) + 190$<br />$h(1.5) = -16(2.25) + 72 + 190$<br />$h(1.5) = -36 + 72 + 190 = 226$ feet.
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