Question 234949: Please help me to solve this problem=
Stu lives on the top floor of an apartment building, and Hal lives on a lower floor. Hal's window is 1/4 as high as Stu's. Stu drops a stone from his window. Three seconds later, Hal drops a stone from his window. The two stones hit the ground simultaneously. How high is Stu's window from the ground?
Thank you for helping me to solve this problem.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! This formula for distance of a free falling object might help.
Stu's on top.
Hal's below.
Hal's window is 1/4 as high as Stu's.
This means that Stu's window is 4 times as high as Hal's.
Stu drops a stone from his window.
3 seconds later, Hal drops a stone from his window.
The 2 stones hit the ground simultaneously.
How high is Stu's window from the ground?
The formula that will help solve this problem is:
D = A * T^2 / 2
where:
D = distance in meters.
A = acceleration of Gravity = 9.8 m/s^2 (meters per second squared).
T = time in seconds.
Stu drops his stone first.
3 seconds later Hal drops his stone.
The distance that Stu's stone has to travel is 4 times the distance that Hal's stone has to travel.
If we let D = the distance that Hal's stone has to travel, then the formula for the distance that Hal's stone traveled becomes:
D = A * T^2 / 2.
Since Stu's traveled 4 times the distance that Hal's stone traveled, and Stu's stone traveled for 3 more seconds than Stu's stone traveled, then the formula for the distance that Stu's stone traveled becomes:
4D = A * (T+3)^2 / 2
We have two formulas that need to be solved simultaneously because the same D and the same T apply to both.
They are:
D = (A * T^2 / 2)
and
4D = (A * (T+3)^2 / 2)
We can substitute for D from the first equation and solve for T in the second equation.
The second equation becomes:
4*A*T^2 / 2 = A*(T+3)^2 / 2
We multiply both sides of the equation by 2 and we divide both sides of the equation by A to get:
4*T^2 = (T+3)^2
We expand the right side of the equation to get:
4*T^2 = T^2 + 6*T + 9
We subtract T^2 + 6*T + 9 from both sides of the equation to get:
4*T^2 - T^2 - 6*T - 9 = 0
We combine like terms to get:
3*T^2 - 6*T - 9 = 0
We divide both sides of this equation by 3 to get:
T^2 - 2*T - 3 = 0
We factor this quadratic equation to get:
(T-3) * (T+1) = 0
We solve for T to get:
T = 3 or T = -1
Since T can't be negative (T represents time), we get:
T = 3
We test this value for T to see if it is good.
Our original equations are:
D = (A * T^2) / 2
and
4D = (A * (T+3)^2) / 2
The first equation becomes D = (A * 3 ^2) / 2 = 9*A/2
the second equation becomes 4D = (A * 6^2) / 2 = 36*A/2
We divide the second equation by 4 to get D = 9*A/2
The equations are identical so the value for T is good.
We now look to solve for D.
We know that A = 9.8 meters per second squared.
Our equations become:
D = (9.8 * 9) / 2 = 44.1 meters
4D = (9.8 * 36) / 2 = 176.4 meters
Stu's window is 176.4 meters from the ground.
this is 4 times the height of Hal's window which is 44.1 meters from the ground, as it should be.
The answer to the question is that Stu's window is 176.4 meters from the ground.
If you look at what has happened, the second by second sequence is shown below, using the formula D = 9.8*T^2:
Results are truncated to the nearest 10th of a second.
after 1 second Stu's stone has traveled 4.9 meters
after 2 seconds Stu's stone has traveled 19.6 meters
after 3 seconds Stu's stone has traveled 44.1 meters
after 4 seconds Stu's stone has traveled 78.4 meters
after 5 seconds Stu's stone has traveled 122.5 meters
after 6 seconds Stu's stone has traveled 176.4 meters.
The figures for Hal are:
after 1 second Hal's stone has traveled 4.9 meters
after 2 seconds Hal's stone has traveled 19.6 meters
after 3 seconds Hal's stone has traveled 44.1 meters
Hal started 3 seconds after Stu.
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