SOLUTION: I really need help with the following problem. Thank you! Science and Medicine: Find the time required for an object to fall to the ground from a building that is 1400 ft. hig

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I really need help with the following problem. Thank you! Science and Medicine: Find the time required for an object to fall to the ground from a building that is 1400 ft. hig      Log On


   

Question 77919: I really need help with the following problem. Thank you!
Science and Medicine: Find the time required for an object to fall to the ground from a building that is 1400 ft. high. Show thew formula and solve.



Found 2 solutions by Earlsdon, bucky:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You can use the free-fall equation: h%28t%29+=+-%281%2F2%29gt%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D where:
g is the constant of acceleration due to gravity = 32 ft/sec^2
v%5B0%5D is the initial velocity of the object which, in this case, is zero.
h%5B0%5D is the initial height of the object above the ground which, in this case is 1400 feet.
t is the time in seconds.
You are looking for the value of t when h = 0 feet. So, putting in the numbers, we get:
h%28t%29+=+-16t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D Setting h(t) = 0 (for zero height) andv%5B0%5D=0
0+=+-16t%5E2%2B0%2B1400 Simplifying this:
16t%5E2+=+1400 Divide both sides by 16.
t%5E2+=+1400%2F16
t%5E2+=+87.5 Take the square root of both side.
t+=+9.35 to the nearest hundredth. Discard the negative answer as time is a positve value.
It will take 9.35 seconds for the object to reach the ground from a height of 1400 feet.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Find the time required for an object to fall to the ground from a building that is 1400 ft. high.
.
The equation for the height of a falling body is (assuming you neglect air resistance
which would slow the rate of fall a little and therefore increase the time that it takes):
.
h%28t%29+=+%28g%2F2%29%2At%5E2+%2B+%28V%5Bo%5D%2At%29+%2B+h%5Bo%5D
.
in which the terms are defined as follows:
.
h(t) = height as it relates to the time t
.
g = acceleration due to gravity. Usually we use a value of -32 ft/sec^2 for g, but it
really depends where you are on the globe. If you are on top of Mount Everest, the value
of gravity is different from if you are way down in Death Valley, California. It's not
a big difference, but it is different. (The minus sign on gravity is just to show that
it acts downward toward the center of the Earth.)
.
t = time elapsed in seconds
.
V%5Bo%5D is the initial velocity. If you throw the object up or if you throw the object
down, V%5Bo%5D will have a value. If you just let go of the object, V%5Bo%5D will be zero. In
this problem the object is just released so V%5Bo%5D=0
.
h%5Bo%5D is the initial height of the object at t = 0 seconds. In this case the initial
height is 1400 ft because the object is 1400 feet above ground at the time of release.
.
Substituting -32 for g, zero for V%5Bo%5D and 1400 for h%5Bo%5D changes the general
equation to one applicable to this problem. With these substitutions the equation
becomes:
.
h%28t%29+=+-16t%5E2+%2B+1400
.
Notice that when t = 0, the equation reduces to h%28t%29+=+1400 which tells you that at
the instant of release the object is at 1400 ft above the ground.
.
How high will the object be when it hits ground? The height above ground at the time of
impact will be zero. Substitute 0 for h%28t%29 in the equation to get:
.
0+=+-16t%5E2+%2B+1400
.
You can eliminate the -16t%5E2 from the right side by adding 16t%5E2 to both sides
to get:
.
16t%5E2+=+1400
.
Divide both sides by 16 to get:
.
t%5E2+=+1400%2F16+=+87.5
.
and solve for t by taking the square root of both sides to get:
.
t+=+sqrt%2887.5%29+=+9.354
.
So the answer is that it takes the object about 9.354 seconds after release to fall 1400 ft.
.
Hope this helps you to understand the problem a little better.