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When an object is dropped from a building, its height above

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When an object is dropped from a building, its height above

by BTGmoderatorLU » Sat Nov 09, 2019 2:26 am
Source: Manhattan Prep

When an object is dropped from a building, its height above the ground is given by the formula \(h = y - gt^2\) where \(y\) is the height of the building in meters, \(g\) is a constant, and \(t\) is the time in seconds since the object was dropped. What is the height of an object dropped from a 100-meter tall building 2 seconds after being dropped?

1) An object is 74 meters above the ground 2 seconds after being dropped from a building that is \(z\) meters tall.
2) The height of an object dropped from a 150-meter tall building is 61.2 meters 3 seconds after being dropped.

The OA is B
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by gmatter2012 » Sun Nov 10, 2019 3:25 am
BTGmoderatorLU wrote:Source: Manhattan Prep

When an object is dropped from a building, its height above the ground is given by the formula \(h = y - gt^2\) where \(y\) is the height of the building in meters, \(g\) is a constant, and \(t\) is the time in seconds since the object was dropped. What is the height of an object dropped from a 100-meter tall building 2 seconds after being dropped?

1) An object is 74 meters above the ground 2 seconds after being dropped from a building that is \(z\) meters tall.
2) The height of an object dropped from a 150-meter tall building is 61.2 meters 3 seconds after being dropped.

The OA is B
To find the height we need the the value of constant g

1) Does not allow us to find g as we do not know the height z . NOT SUFF.
2) Since we have h, y and t we can find g. This will allow us to to find the h as asked in the stem.
SUFF.

Ans- B
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by deloitte247 » Thu Nov 14, 2019 11:26 pm
$$h=gl^2\ where\ h=height\ above\ the\ ground$$
$$y=height\ of\ the\ building=100m$$
g = constant
t = time = 2s $$h=y-gt^2$$
$$h=100-g\left(2\right)^2\ =\ 100-4g\ \ ---\ eqn\left(1\right)$$
So, what is the height of an object dropped from a 100-meter tall building 2 seconds after being dropped?
Statement 1: An object is 74m above the ground 2 seconds after being dropped from a building that is 'z' meters tall.
$$h=y-gt^2$$
$$74=z-g\left(2\right)^2$$
$$74=z-4g$$
Variable 'z' and 'g' are unknown, hence, statement 1 is NOT SUFFICIENT.

Statement 2: The height of an object dropped from a 150m tall building is 61.2 meters 3 seconds after being dropped.
$$h=y-gt^2$$
$$61.2=150-g\left(3\right)^2$$
$$9g=88.8$$
$$g=\frac{88.8}{9}=9.87\ \frac{m}{s}$$
Now, substitute the value of 'g' in equation (1) from the question stem;
h = 100 - 4g
h = 100 - (4 * 9.87)
h = 100 - 39.48
h = 60.52m
Statement 2 alone is SUFFICIENT. Therefore, option B is the correct answer.
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