One object is thrown straight up at 54.9 m/s from the ground at the base of a building with is 157.9 meters tall. At the same time the object is thrown up at the bottom of the building, another object is thrown straight down at the top of the building. If both objects are to collide at a height of 39 meters above the ground while the object launched from the bottom is still going up (has a positive velocity), how fast, in m/s, does the object at the top of the building need to be thrown downwards? In other words, what is the magnitude of the velocity of the object at the top of the building?
Hint: find the time it takes for the object launched from the ground to reach the height of 39 meters by first using an appropriate equation to find its velocity at that point, and then use a simpler equation to find the time needed to get to that velocity (the quadratic equation is not needed for this method). The second object must make it to the same height in the same time, so that time can be used to analyze the second object.
The First Object:
h = Vo*t + 0.5g*t^2 = 39 m.
54.9*t - 4.9*t^2 = 39
Use Quadratic formula.
t = 0.762 s.
The 2nd. Object:
h = Vo*t + 0.5g*t^2 = 157.9-39 = 118.9 m
Vo*0.762 + 4.9*0.762^2 = 118.9
0.762Vo + 2.845 = 118.9
0.762Vo = 118.9 - 2.845 = 116.05
Vo = 152.3 m/s.
To find the magnitude of the velocity of the object at the top of the building, we need to find the time it takes for the object launched from the ground to reach a height of 39 meters.
First, let's find the time it takes for the object launched from the ground to reach a height of 39 meters. We can use the following equation:
h = vi*t + (1/2)*a*t^2
Where:
h = height (39 meters)
vi = initial velocity (54.9 m/s)
a = acceleration (we will assume it is -9.8 m/s^2 since the object is moving in the opposite direction to the positive direction)
Plugging in the values, we have:
39 = 54.9*t + (1/2)*(-9.8)*t^2
Simplifying, we have the quadratic equation:
-4.9*t^2 + 54.9*t - 39 = 0
Now, we can solve this quadratic equation to find the time it takes for the object to reach a height of 39 meters. Alternatively, we can use a different method without using the quadratic equation.
We can use the equation v = vi + at to find the final velocity of the object when it reaches a height of 39 meters. We know that the final velocity at that point is 0 (since the object reaches the highest point and starts to come down) and the initial velocity is 54.9 m/s. The acceleration a is -9.8 m/s^2.
0 = 54.9 + (-9.8)*t
Solving for t, we get:
t = 54.9 / 9.8
t ≈ 5.6 seconds
So, it takes approximately 5.6 seconds for the object launched from the ground to reach a height of 39 meters.
Now, since the second object must make it to the same height in the same time, we can use this time to analyze the second object.
The object at the top of the building starts from rest and accelerates downwards due to gravity. We need to find the magnitude of the velocity it needs to be thrown downwards so that it collides with the other object at a height of 39 meters.
Using the same formula v = vi + at, we have:
0 = vi - 9.8*t
Since the initial velocity vi is 0 (the object starts from rest), we can solve for the magnitude of the velocity vi at the top of the building:
vi = 9.8*t
Substituting the value of t we found earlier, we have:
vi = 9.8 * 5.6
vi ≈ 54.9 m/s
Therefore, the magnitude of the velocity of the object at the top of the building needs to be approximately 54.9 m/s downwards in order to collide with the other object at a height of 39 meters.