Question

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.

Answer 1

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.