# Water Rocket Lab Report

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A model rocket is fired vertically from an elevated launch pad at an initial height of 4 feet at an initial velocity of 350 feet per second. The gravitational force of Earth is pulling the rocket down at a rate of 16 feet per second per second. The formula for, s(t), for the function that models the height of the water balloon at time t, in seconds, is s(t)= -16t2 + 350t + 4. We can determine how long it takes for the rocket to land from the graph where the function intercepts the x-axis because this is the value where the height of the rocket is 0 feet. From the graph, we can see that the rocket lands in 21.886 seconds. It would not make sense for there to be a negative x-value, or a negative y-value since one cannot go back in time, and the rocket cannot go below the …show more content…
This means for every second between 4.5 and 6 seconds, on average, the height increases by 182 feet. From the interval from 6.5 seconds to 7 seconds, I found the average rate of change is 135 feet per seconds. This means for every second between 6.5 and 7 seconds, on average, the height increases by 135 feet. You can make equations with these average rates of change too using the average rate of change and a coordinate you used to find the average rate of change and plug it in to the point-slope equation of (y-y0)=m(x-x0). The equation for the interval between 4.5 seconds and 6 seconds is (y-1255)=182(x-4.5). There are many intervals where the average velocity of the rocket is 0. From 2 seconds to 19.875 seconds is an example of one of these intervals where the average velocity of the rocket is 0 feet per second. This tells you that within the interval between 2 and 19.875 seconds, on average, for every second that passes, the height of the rocket increases by 0 feet. This also tells you that at 2 seconds, the rocket has the same position going up as the rocket at 19.875 seconds going

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