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?
To find the initial distance between the two objects, we need to consider the time it takes for both objects to hit the ground.
First, let's find the time it takes for the object released from rest to fall 20.0 m. We can use the equation of motion for free fall:
h = (1/2) * g * t^2
Where:
h is the height (20.0 m)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time taken
Rearranging the equation, we have:
t^2 = (2 * h) / g
Substituting the values into the equation:
t^2 = (2 * 20.0) / 9.8
t^2 ≈ 4.08
Taking the square root of both sides, we find:
t ≈ 2.02 s
Now, let's find the time it takes for the horizontally shot object to hit the ground. Since there is no vertical acceleration, the time can be calculated solely based on the vertical motion of the object released from rest. Both objects hit the ground at the same time, so the time it takes for the horizontally shot object to travel the initial distance needs to be the same as the time calculated above.
We can use the equation of motion, s = v * t, where s is the distance and v is the velocity:
s = v * t
Substituting the values into the equation:
s = 2.5 m/s * 2.02 s
s ≈ 5.05 m
Therefore, the initial distance between the two objects was approximately 5.05 meters.
To solve this problem, we can break it down into two parts: the vertical motion and the horizontal motion.
1) Vertical motion:
Let's assume the two objects collide after a time t. We can use the equation of motion for free fall to find the time it takes for an object to fall a certain distance.
The equation of motion for free fall is: h = 1/2 * g * t^2, where h is the vertical distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
For the object released from rest:
The vertical distance it travels can be calculated as: h1 = 1/2 * g * t^2
For the object shot horizontally:
When an object is shot horizontally, its vertical motion is the same as that of an object released from rest. Thus, the vertical distance it travels can also be calculated as: h2 = 1/2 * g * t^2
Since the two objects collide after falling 20.0 m, we can set h1 + h2 = 20.0 m:
1/2 * g * t^2 + 1/2 * g * t^2 = 20.0 m
Simplifying the equation, we have: g * t^2 = 40.0 m
2) Horizontal motion:
The horizontal motion of the object shot horizontally is uniform with an initial speed of 2.5 m/s. We can use the formula: d = v * t, where d is the horizontal distance, v is the initial velocity, and t is the time.
Since both objects are released at the same time, the time taken is the same for both objects. Therefore, the horizontal distances traveled by the two objects are equal.
Let's assume the horizontal distance traveled by both objects is d.
According to the information provided, the horizontal distance traveled by the object shot horizontally is d = 2.5 * t.
To find the initial distance between the objects, we need to find the time it takes for the objects to collide.
Substituting d = 2.5 * t and g * t^2 = 40.0 m into the equation: d = 2.5 * t, we have:
2.5 * t = 40.0
Solving for t, we get: t = 16 s
Now, we can substitute the value of t into the equation d = 2.5 * t to find the initial distance:
d = 2.5 * 16 = 40 m
Therefore, the initial distance between the objects was 40 meters.