A certain freely falling object requires 1.50 seconds to travel the las 30.0 meters before it hits th ground. From what height above the ground did it fall?
To determine the initial height from which the object fell, we can use the equation of motion for objects in free fall:
h = (1/2) * g * t^2
where:
- h is the height
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken to fall
Given:
- Distance traveled in the last 30.0 meters
- Time taken is 1.50 seconds
We can rearrange the equation to solve for the initial height:
h = (1/2) * g * t^2
h = (1/2) * 9.8 m/s^2 * (1.50 s)^2
h = (1/2) * 9.8 m/s^2 * 2.25 s^2
h = 1/2 * 9.8 m/s^2 * 2.25 s^2
h = 1/2 * 9.8 m/s^2 * 2.25 s^2
h = 11.025 m
Therefore, the object fell from a height of 11.025 meters above the ground.
To determine the height from which the object fell, we need to use the kinematic equation for free-falling objects. This equation relates the distance an object falls (h), the initial velocity (v₀), gravitational acceleration (g), and time (t):
h = v₀t + (1/2)gt²
In this case, the initial velocity is zero because the object begins its fall from rest. The acceleration due to gravity (g) is approximately 9.8 m/s². The time (t) is given as 1.50 seconds, and we are looking for the height (h).
So, we can rewrite the equation as:
h = (1/2)gt²
Substituting the given values:
h = (1/2)(9.8 m/s²)(1.50 s)²
h = (1/2)(9.8 m/s²)(2.25 s²)
h = 11.025 m
Therefore, the object fell from a height of 11.025 meters above the ground.
V = 30m/1.5s = 20 m/s = Final velocity.
h = (V^2-Vo^2)/2g.
h = (400-0)/19.6 = 20.41 m.