Answer:
a) The height of the small object 3 seconds after being launched is 2304 feet.
b) The small object ascends 128 feet between 5 seconds and 7 seconds.
c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.
d) The object will take 21 seconds to hit the ground.
Step-by-step explanation:
The correct formula for the height of the small object is:
[tex]h(t) = -16\cdot t^{2}+256\cdot t+1680[/tex] (1)
Where:
[tex]h[/tex] - Height above the ground, measured in feet.
[tex]t[/tex] - Time, measured in seconds.
a) The height of the small object at given time is found by evaluating the function:
[tex]h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680[/tex]
[tex]h(3\,s) = 2304\,ft[/tex]
The height of the small object 3 seconds after being launched is 2304 feet.
b) First, we evaluate the function at [tex]t = 5\,s[/tex] and [tex]t = 7\,s[/tex]:
[tex]h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680[/tex]
[tex]h(5\,s) = 2560\,s[/tex]
[tex]h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680[/tex]
[tex]h(7\,s) = 2688\,s[/tex]
We notice that the small object ascends in the given interval.
[tex]\Delta h = h(7\,s)-h(5\,s)[/tex]
[tex]\Delta h = 128\,ft[/tex]
The small object ascends 128 feet between 5 seconds and 7 seconds.
c) If we know that [tex]h = 2640\,ft[/tex], then (1) is reduced into this second-order polynomial:
[tex]-16\cdot t^{2}+256\cdot t-960=0[/tex] (2)
All roots of the resulting equation come from the Quadratic Formula:
[tex]t_{1} = 10\,s[/tex] and [tex]t_{2}= 6\,s[/tex]
The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.
d) If we know that [tex]h = 0\,ft[/tex], then (1) is reduced into this second-order polynomial:
[tex]-16\cdot t^{2}+256\cdot t +1680 = 0[/tex] (3)
All roots of the resulting equation come from the Quadratic Formula:
[tex]t_{1} = 21\,s[/tex] and [tex]t_{2} = -5\,s[/tex]
Just the first root offers a solution that is physically reasonable.
The object will take 21 seconds to hit the ground.
