A girl throws a stone form the top of a building with an initial velocity of 20.0 m/s upward. The building is 50.0 m high and the stone just misses the edge of the roof on its way down. Draw a v vs t graph, find time needed for stone to reach max height, find max height, time needed for the stone to return to the level of the girl, find velocity of the stone when it returns to the level of the girl and find the time for the stone to reach the ground..
You are going to have to show some work of your own.
V = 20 - 9.8 t
max height is reached when V = 0
Height above ground =
Y = 50 + 20t - 4.9 t^2
Y = 0 when it touches the ground. Solve for that t.
a person throws an apple downward from a lift the man observes doesn't fall down why?
To begin, let's analyze the motion of the stone.
1. Drawing the v vs t graph:
Since the initial velocity is 20.0 m/s upward, the stone starts with a positive velocity. As it moves upwards, the velocity decreases until it reaches its maximum height, at which point the velocity becomes zero. As it falls back down, the velocity increases in the negative direction. The graph would look like this:
|
20 | ^
| /
| /
| /
| /
0 |----------------------/------ t
|
|----------------------|
2. Time needed for the stone to reach maximum height:
We know that the initial velocity is 20.0 m/s and the final velocity at maximum height is 0 m/s. Due to symmetrical motion, it will take the same time to reach the maximum height as it does to return to the point of release. Let's use the kinematic equation:
vf = vi + at
Since vf = 0 m/s, vi = 20.0 m/s, and a = -9.8 m/s² (due to gravity), we can solve for time:
0 = 20.0 - 9.8t
Simplifying, we get:
9.8t = 20.0
t ≈ 2.04 seconds
Therefore, it takes approximately 2.04 seconds for the stone to reach its maximum height.
3. Maximum height:
To find the maximum height, we can use the kinematic equation:
vf² = vi² + 2ad
Since we want to find the height, the final velocity at maximum height is 0 m/s, the initial velocity is 20.0 m/s, and the acceleration is -9.8 m/s², we can substitute these values:
0 = (20.0)² + 2(-9.8)d
Solving for d (maximum height):
400 = -19.6d
d ≈ -20.4 meters
The negative sign indicates that the stone is below the starting point, so we take the magnitude and consider the maximum height as approximately 20.4 meters above the ground.
4. Time needed for the stone to return to the level of the girl:
As mentioned earlier, it takes the same amount of time for the stone to reach maximum height as it does to return to the point of release. Therefore, it will also take approximately 2.04 seconds for the stone to return to the level of the girl.
5. Velocity of the stone when it returns to the level of the girl:
To find the velocity, we can use the equation vf = vi + at.
Since we want to find the final velocity, vi = 0 m/s, a = -9.8 m/s², and the time is approximately 2.04 seconds, we can substitute these values:
vf = 0 - 9.8(2.04)
vf ≈ -19.99 m/s
Therefore, the velocity of the stone when it returns to the level of the girl is approximately -19.99 m/s, pointing downward.
6. Time for the stone to reach the ground:
The time for the stone to reach the ground can be calculated using the equation vf = vi + at.
Since the initial velocity vi = 20.0 m/s, the final velocity vf is when the stone hits the ground, and a = 9.8 m/s² (taking the positive value due to the downward motion), we can solve for time:
vf = 20.0 + 9.8t
0 = 20.0 + 9.8t
Solving for t:
9.8t = -20.0
t ≈ -2.04 seconds
The negative sign indicates that the time is in the opposite direction of our initial reference point. We can take the magnitude and consider the time for the stone to reach the ground as approximately 2.04 seconds.
Thus, the stone takes approximately 2.04 seconds to reach the ground.