A body is dropped from the rooftop of the building.The body reaches the ground at t= 4.05 from the point of release.how tall is the building?what is the speed of the body just before it reaches the ground?
The symbol t is usually not used to denote distance. It is used for time. Did you copy the problem correctly? What are the dimensions that go with the number 4.05?
i copied the problem correctly sir :)
The problem would make sense if it said the body reaches the ground 4.05 seconds after it is dropped. If it is really the way you said, with no dimensions after the number 4.05, you should be learning physics somewhere else.
To calculate the height of the building, we can use the formula for the free fall motion:
h = (1/2) * g * t^2
where:
h = height of the building
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time it takes for the body to reach the ground (4.05 seconds)
Substituting the given values into the formula:
h = (1/2) * 9.8 * (4.05^2)
h ≈ 79.59 meters
Therefore, the height of the building is approximately 79.59 meters.
To calculate the speed of the body just before it reaches the ground, we can use the formula for the final velocity in free fall motion:
v = g * t
Substituting the values:
v = 9.8 * 4.05
v ≈ 39.69 m/s
Therefore, the speed of the body just before it reaches the ground is approximately 39.69 m/s.