Question

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)

Answer 1

Determine the value of v, rounded to the nearest whole number.

⇒ 56

Which quadratic equation models the situation correctly?

h(t) = –16t2 + 56t + 6.5

Explanation:

on edgy

Answer 2

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