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.
the average velocity during the 2m is 2/.22sec. That average velocity occurs at the center, or 1 m off the ground.
Now, at the top, Vi=0
V(1m)=Vi+gt
solve for t, time to fall all but the last meter.
hf=ho+Vi*t-4.9t^2
1=ho-4.9t^2 solve for ho using the time above.
To determine the height of the building, we can use the equation of motion for vertical motion:
h = (1/2) * g * t^2
where:
h is the height of the building
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken for the sandbag to fall (0.22 s)
First, we need to calculate the velocity of the sandbag when it hits the ground, using the equation:
v = g * t
Substituting the given values, we have:
v = 9.8 m/s^2 * 0.22 s
v ≈ 2.156 m/s
Now, we can calculate the height of the building using the equation of motion:
h = (1/2) * g * t^2
Substituting the values:
h = (1/2) * 9.8 m/s^2 * (0.22 s)^2
h ≈ 0.21924 m
Therefore, the height of the building is approximately 0.21924 meters.