2) Two objects are initially the same height above the ground. Simultaneously, one is released from rest and the other is shot horizontally with an initial speed of 2.5 m/s. The two objects collide after falling 20.0 m. How far apart were the objects initially?
10.2m
fewqre
To solve this problem, we need to break it down into two parts: vertical motion and horizontal motion. Let's start with the vertical motion:
Let's assume that the time it takes for both objects to reach the ground is 't' seconds. We can use the equation of motion for vertical motion:
s = ut + 0.5gt^2
Where:
s = displacement (20.0 m)
u = initial velocity (0 m/s for the dropped object, since it's released from rest)
g = acceleration due to gravity (-9.8 m/s^2)
For the object that is dropped, the equation becomes:
20.0 = 0*t + 0.5*(-9.8)*t^2
Simplifying this equation gives:
4.9t^2 = 20.0
Dividing both sides by 4.9, we get:
t^2 = 4.08
Taking the square root of both sides, we find:
t = 2.02 seconds (approx.)
Now, let's move on to the horizontal motion:
For the object that is shot horizontally, we know that its initial velocity in the horizontal direction is 2.5 m/s. Since there is no acceleration in the horizontal direction, the horizontal displacement can be found using the equation:
s = ut
Where:
s = horizontal displacement
u = initial velocity in the horizontal direction
t = time taken to collide (2.02 seconds)
Plugging in the values, we get:
s = 2.5*2.02
s = 5.05 meters
So, the objects were initially 5.05 meters apart horizontally.