A stone is dropped from the top of a tall building and at the same time another stone is thrown vertically upward from the ground level with a velocity of 20m/s the stones meet exectity 2s after their release how tall is the building
Since the height of the falling stone is the same as that of the projected stone after 2 seconds,
h-4.9t^2 = 20t-4.9t^2
h = 20t = 40
Why did the stone take so long to meet the other stone? It sounds like they both needed a serious lesson in punctuality. As for the height of the building, since the stones meet exactly 2 seconds after their release, we can calculate the height using the equation for the vertical motion of an object:
H = (1/2)gt^2
Where H is the height, t is the time, and g is the acceleration due to gravity. In this case, the stone dropped from the top of the building, so it experienced free fall. Thus, we can approximate g as 9.8 m/s^2. Plugging in the values, we get:
H = (1/2)(9.8)(2)^2
H = 19.6 meters
So, the height of the building is approximately 19.6 meters. Just remember, next time the stones decide to have a meeting, they should really work on their timing!
To find the height of the building, we need to use the equations of motion for both stones. Let's consider the motion of each stone separately:
For the stone being dropped:
1. Using the equation of motion for displacement (s) in free fall: s = ut + (1/2)at^2
where u = initial velocity = 0, a = acceleration due to gravity = 9.8 m/s^2, and t = time = 2 s.
Substituting the values, we get:
s1 = 0 + (1/2)(9.8)(2)^2
= (1/2)(9.8)(4)
= 19.6 m
For the stone thrown upwards:
1. Using the equation of motion for displacement (s) when thrown upwards: s = ut + (1/2)at^2
where u = initial velocity = 20 m/s, a = acceleration due to gravity = -9.8 m/s^2 (negative as it opposes motion), and t = time = 2 s.
Substituting the values, we get:
s2 = (20)(2) + (1/2)(-9.8)(2)^2
= 40 - 19.6
= 20.4 m
Since the stones meet exactly 2 seconds after their release, the height of the building is equal to the sum of the displacements of both stones:
Total height of the building = s1 + s2
= 19.6 + 20.4
= 40 meters
Therefore, the height of the building is 40 meters.
To determine the height of the building, we need to find the time it takes for the stone dropped from the building to reach the ground. Since both stones meet exactly 2 seconds after their release, we know that the stone thrown vertically upward took 2 seconds to reach the meeting point.
Now, let's break down the problem further. We have two stones, one thrown upward and one dropped from the top of the building. Let's analyze the motion of each stone separately:
1. Stone thrown upward:
- Initial velocity: 20 m/s upward (positive velocity)
- Time taken to reach the meeting point: 2 seconds
- We can use the equation: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, t is the time taken, and a is the acceleration.
- Since the stone is thrown vertically upward, the acceleration is due to gravity, which is approximately -9.8 m/s^2 (negative value because it acts downward).
- Plugging in the values:
s = (20 m/s)(2 s) + (1/2)(-9.8 m/s^2)(2 s)^2
s = 40 m - 19.6 m
s = 20.4 m upward (displacement of the stone thrown upward)
2. Stone dropped from the top of the building:
- Initial velocity: 0 m/s (stone is dropped, so the initial velocity is zero)
- We need to find the time it takes to reach the ground, which is the meeting point with the stone thrown upward.
- The equation for the vertical distance (height) traveled by a falling object is: s = (1/2)gt^2, where g is the acceleration due to gravity.
- Plugging in the values:
20.4 m = (1/2)(-9.8 m/s^2)t^2
40.8 m = -4.9 m/s^2 * t^2
t^2 = -40.8 m / -4.9 m/s^2
t^2 = 8.33 s^2
t ≈ √8.33 s
t ≈ 2.89 s (time taken by the stone dropped from the building to reach the ground)
Since both stones meet after 2 seconds, and the stone dropped from the building takes an additional 2.89 seconds to reach the ground, the height of the building can be calculated by multiplying the time taken by the stone dropped from the building by the average velocity during that time:
Height of the building = (2.89 s) * (20 m/s)
Height of the building ≈ 57.8 m
Therefore, the height of the building is approximately 57.8 meters.