An object is dropped from a height H above the ground. This free-falling object requires 0.904 s to travel the last 34 m before hitting the ground.
From what height H above the ground did
the object fall? The acceleration of gravity is 9.8 m/s^2.
Answer in units of m
i understand that t= .904s
distance is 34m
and a = 9.8m/s^2
But I'm not sure what else to do here?
please someone help!!
45.47m/s
To determine the height H from which the object fell, we can use the equations of motion for a freely falling object.
The equation we can use in this case is: h = ut + (1/2)at^2
Where:
- h is the height the object fell from (which is what we want to find)
- u is the initial velocity (in this case, the object was dropped, so u = 0)
- a is the acceleration due to gravity (9.8 m/s^2)
- t is the time it took to travel the last 34 m (0.904 s)
In this case, since the object was dropped, the initial velocity u is 0.
So the equation becomes: h = 0 + (1/2)(9.8)(0.904)^2
Simplifying this equation, we have: h = 0.5(9.8)(0.817216)
Calculating this, h ≈ 3.978 m
Therefore, the object fell from a height of approximately 3.978 m above the ground.
To find the height H from which the object fell, we need to use the equations of motion for an object in freefall. The equation that relates distance, time, and acceleration is:
Distance (d) = Initial Velocity (u) x Time (t) + 0.5 x Acceleration (a) x Time (t)^2
In this case, the initial velocity (u) of the object is 0, as it was dropped vertically downward. We are given the time (t) and the distance (d).
Using the equation above, we can rearrange it to find the initial height (H) from which the object fell. Let's go through the steps:
1. Rearrange the equation:
0.5 x a x t^2 = d
2. Plug in the known values:
0.5 x 9.8 m/s^2 x (0.904 s)^2 = 34 m
3. Simplify the equation:
4.4 x (0.904 s)^2 = 34 m
4. Solve for the initial height, H:
H = (34 m) - 4.4 x (0.904 s)^2
H = 34 m - 3.92 m
H = 30.08 m
Therefore, the object fell from a height H = 30.08 m above the ground.