a projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. the…

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

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

Answer

Answer:

C. 226 feet

Explanation:

Step1: Encontrar el tiempo en el máximo

La ecuación de la altura es $h(t)=-16t^{2}+48t + 190$, que es una parábola de la forma $y = ax^{2}+bx + c$ con $a=-16$, $b = 48$ y $c = 190$. El tiempo $t$ en el vértice (máximo) de una parábola se da por $t=-\frac{b}{2a}$. $t=-\frac{48}{2\times(-16)}=\frac{48}{32}=\frac{3}{2}$ segundos.

Step2: Calcular la altura máxima

Sustituir $t = \frac{3}{2}$ en la ecuación $h(t)$. $h(\frac{3}{2})=-16\times(\frac{3}{2})^{2}+48\times\frac{3}{2}+190$. $h(\frac{3}{2})=-16\times\frac{9}{4}+72 + 190$. $h(\frac{3}{2})=-36+72 + 190$. $h(\frac{3}{2})=36+190=226$ pies.