Question

this question has two parts. first, answer part a. then, answer part b. part a volleyball camden hits a volleyball straight up at velocity of 10.78 meters per second from a height of 1.9 meters. part a find and interpret the axis of symmetry. x = select choice ; points equidistant from the axis of symmetry represent times when the height of the volleyball will be select choice. part b part b find and interpret the vertex. vertex: select choice ; the volleyball reaches a select choice height of select choice meters select choice seconds after it is hit.

Explanation:

Step1: Ecuación de movimiento

La ecuación de la altura \( h(t) \) de un objeto lanzado verticalmente se da por \( h(t)=-4.9t^{2}+v_{0}t + h_{0} \), donde \( v_{0} = 10.78\ m/s \) (velocidad inicial) y \( h_{0}=1.9\ m \) (altura inicial). Entonces, \( h(t)=-4.9t^{2}+10.78t + 1.9 \).

Step2: Eje de simetría de una parábola

Para una parábola \( y = ax^{2}+bx + c \), el eje de simetría se da por \( x=-\frac{b}{2a} \). En \( h(t)=-4.9t^{2}+10.78t + 1.9 \), \( a=-4.9 \) y \( b = 10.78 \). Entonces, \( t=-\frac{10.78}{2\times(-4.9)}=\frac{10.78}{9.8}=1.1 \).

Step3: Vértice de la parábola

Sustituimos \( t = 1.1 \) en \( h(t) \): \( h(1.1)=-4.9\times(1.1)^{2}+10.78\times1.1 + 1.9=-4.9\times1.21+11.858+1.9=-5.929+11.858+1.9 = 7.829\approx7.83 \).

Answer:

Part A: \( x = 1.1 \); points equidistant from the axis of symmetry represent times when the height of the volleyball will be the same.
Part B: Vertex: \( (1.1,7.83) \); the volleyball reaches a maximum height of \( 7.83 \) meters \( 1.1 \) seconds after it is hit.