Question

At the local downtown 4th of July fireworks celebration, the fireworks are shot by remote control into the air from a pit in the ground that is 12 feet below the earth's surface. Find an equation that models the height of an aerial bomb above the ground (in feet) t seconds after it is shot upwards with an initial velocity of 80 ft/sec.

A. h(t) = 12 + 80t - 16t^2
B. h(t) = 12 - 80t - 16t^2
C. h(t) = 12 + 80t + 16t^2
D. h(t) = 12 - 80t + 16t^2

Answer 1

The equation that models the height of an aerial bomb above the ground (in feet) t seconds after it is shot upwards with an initial velocity of 80 ft/sec is h(t) = 12 + 80t - 16t².

Step-by-step explanation:

The equation that models the height of an aerial bomb above the ground (in feet) t seconds after it is shot upwards with an initial velocity of 80 ft/sec is:

h(t) = 12 + 80t - 16t²

This equation represents the height of the bomb as a function of time. The initial height of the bomb is 12 feet, and it increases linearly with time at a rate of 80 feet per second. However, the height also decreases quadratically with time, due to the effect of gravity. The term -16t² represents the gravitational acceleration pulling the bomb downwards.