The function H(t) = −16t2 + 48t + 12 shows the height H(t), in feet, of a cannon ball after t seconds. A second cannon ball moves in the air along a path represented by g(t) = 10 + 15.2t, where g(t) is the height, in feet, of the object from the ground at time t seconds. Part A: Create a table using integers 0 through 3 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points) Part B: Explain what the solution from Part A means in the context of the problem. (4 points)
Given two functions: h(t)=-16t^2+48t+12 and g(t)=10+15.2t A] The table of both functions from 0 to 3 will be: h(t)=-16t^2+48t+12 t      0      1       2      3 h(t)    12     44      44     12
g(t)=10+15.2t
t      0     1       2       3 g(t)   10    25.2    40.4     55.6
the point in which h(t)=g(t) will be given by: -16t^2+48t+12=10+15.2t forming quadratic equation we get: -16t^2+32.8t+12=0 solving the above quadratic equation using the formula we get: t=-0.32 or t=2.4 therefore we conclude that they only met once, at point t=2.4 sec therefore they met between points t=-0.32 and t=2.4
