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