A bag of sand dropped by a would-be assassin from the roof of a building just misses Tough Tony, a gangster 2-m tall. The missile traverses the height of Tough Tony in 0.22 s, landing with a thud at his feet. How high was the building? Ignore friction.
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To find the height of the building, we can use the equation of motion in vertical direction:
h = u*t + (1/2) * g * t^2
Where:
h = height of the building
u = initial vertical velocity of the sandbag (which is 0 as it was dropped)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken for the sandbag to fall (0.22 s)
Let's substitute the known values into the equation:
h = 0 * 0.22 + (1/2) * (-9.8) * (0.22)^2
h = 0 + (-4.9) * 0.0484
h = -0.23716
The negative sign indicates that the height is below the starting point. However, since we are assuming a positive direction for height, we can consider the magnitude of the height only. Hence, the height of the building is approximately 0.24 meters.
To determine the height of the building, we can use the equations of motion in freefall.
Let's denote the initial velocity of the bag of sand as u (which is 0, as it's dropped from rest), the final velocity as v, the time taken as t, and the height of the building as h.
First, we can find the acceleration (a) experienced by the bag of sand during its fall using the equation:
v = u + at
Since the bag of sand is dropped from rest, u = 0, and v is the final velocity when it reaches the ground, which is also 0. Thus, the equation becomes:
0 = 0 + at
This implies that the acceleration, in this case, is 0 m/s^2. This means the bag of sand falls with constant velocity.
Next, we can use the equation:
h = ut + (1/2)at^2
Plugging in the known values:
h = 0 + (1/2)(0)(0.22)^2
Since the acceleration is 0, this term becomes 0. The equation simplifies to:
h = 0
Therefore, the height of the building is 0 meters. This means the bag of sand was dropped from the same height as Tough Tony, and it just missed him by landing at his feet.