When will they pass each other>

[rlmurra2]

An object is thrown downward w/ initial speed (t=0) of 10 m/s from height 60 m above the ground. At the same instant (t=0) a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?

Ok ... I know the initial velocities, initial time, initial position...Im stuck, this should be really easy if I just get a big hint. I'm going to go work on this and try to figure it out.

 

[GCT]

Well, you know the position value will be the same, translate each situation in terms of distance.

 

[thepatientmental]

Based on the given information, it is possible to calculate the time it takes for each object to reach the same height. Once the time is known, the position of each object can be calculated and compared to determine when they will pass each other. The equation for the position of an object in free fall is s = ut + 1/2at^2, where s is the position, u is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. For the first object, s = 60 m, u = 10 m/s, and a = 9.8 m/s^2. Solving for t gives t = 3.46 seconds. For the second object, s = 0 m, u = 40 m/s, and a = -9.8 m/s^2 (since it is moving in the opposite direction). Solving for t gives t = 4.08 seconds. Therefore, the objects will pass each other at a time between 3.46 and 4.08 seconds. To find the height at which they pass each other, we can use the equation for the position of the first object and substitute t = 3.46 seconds, giving s = 60 + 10(3.46) + 1/2(9.8)(3.46)^2 = 93.6 m. Therefore, the objects will pass each other at a height of 93.6 meters above the ground.