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projectile a projectile is launched at a velocity of 64 feet per second…

Question

projectile a projectile is launched at a velocity of 64 feet per second straight upwards from a height of 3 feet. the height of the projectile after t seconds is given by the function h(t) = -32t² + 64t + 3. find and interpret the vertex of the graph. the projectile reaches a height of select choice feet select choice second(s) after launch.

Explanation:

Step1: Encontrar el tiempo del vértice

Para una función cuadrática \( h(t) = at^2 + bt + c \), el tiempo \( t \) del vértice se calcula con \( t = -\frac{b}{2a} \). Aquí, \( a = -32 \) y \( b = 64 \), así que:
\( t = -\frac{64}{2(-32)} = -\frac{64}{-64} = 1 \)

Step2: Encontrar la altura del vértice

Sustituir \( t = 1 \) en \( h(t) \):
\( h(1) = -32(1)^2 + 64(1) + 3 = -32 + 64 + 3 = 35 \)

Step3: Interpretar el vértice

El vértice es \( (1, 35) \), lo que significa que después de 1 segundo, el proyectil alcanza su altura máxima de 35 pies.

Answer:

The projectile reaches a height of 35 feet 1 second(s) after launch.