Question

We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance of the object from the ground. Exactly seconds after we threw the object, it’s height, (measured in feet) is
ℎ= -16^2 + 256 + 1680

a.) Find ℎ when = 3.

b.) How much does the object travel during the two seconds between 5 seconds and 7 seconds?

c.) How long does it take for the object to reach a height of 2640 ft?

d.) How long does it take for the object to hit the ground?

Answer 1

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:

$h(t) = -16\cdot t^{2}+256\cdot t+1680$ (1)

Where:

$h$ - Height above the ground, measured in feet.

$t$ - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

$h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680$

$h(3\,s) = 2304\,ft$

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at $t = 5\,s$ and $t = 7\,s$:

$h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680$

$h(5\,s) = 2560\,s$

$h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680$

$h(7\,s) = 2688\,s$

We notice that the small object ascends in the given interval.

$\Delta h = h(7\,s)-h(5\,s)$

$\Delta h = 128\,ft$

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that $h = 2640\,ft$, then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t-960=0$ (2)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 10\,s$ and $t_{2}= 6\,s$

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that $h = 0\,ft$, then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t +1680 = 0$ (3)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 21\,s$ and $t_{2} = -5\,s$

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.