Question

a ball is thrown upward from an initial height of 40 feet with an initial velocity of 12 feet per second. after t seconds, the height of the ball can be described by the polynomial $-16r^{2}+12t + 40$.
step 1 of 2: what is the height of the ball after 1 second?

Explanation:

Step1: Substitute \( t = 1 \) into the polynomial

Given the polynomial \( h(t)=-16t^{2}+12t + 40 \), when \( t = 1 \), we have \( h(1)=-16\times(1)^{2}+12\times(1)+40 \).

Step2: Calculate each term

First, calculate \( -16\times(1)^{2}=-16\times1=-16 \). Second, calculate \( 12\times(1) = 12 \).

Step3: Sum up the terms

\( h(1)=-16 + 12+40=(-16+12)+40=-4 + 40=36 \).

Answer:

36