An object falls a distance h from rest. If it travels 0.43h in the last 1.00 s, find (a) the time and (b) the height of its fall.
To solve this problem, we can use the equations of motion for an object in free fall.
Let's analyze the information given in the problem:
We are given:
- Distance traveled in the last 1.00 second = 0.43h
We need to find:
(a) The time it takes for the object to fall
(b) The height of its fall
(a) To find the time it takes for the object to fall, we can use the equation of motion:
h = (1/2)gt^2
where:
h is the height of the fall,
g is the acceleration due to gravity (which is approximately 9.8 m/s² on Earth),
t is the time taken to fall.
Since we are given that the distance traveled in the last 1.00 second is 0.43h, we can set up the following equation:
0.43h = (1/2)gt^2
Now let's solve for t:
Multiply both sides of the equation by 2:
0.86h = gt^2
Divide both sides of the equation by g:
0.86h/g = t^2
Take the square root of both sides:
√(0.86h/g) = t
So, we have found the expression for time t.
(b) To find the height of the fall, we can use the equation of motion:
h = (1/2)gt^2
Substituting the value of t found in part (a) into the equation, we get:
h = (1/2)g(√(0.86h/g))^2
Simplifying the equation:
h = (1/2)g(0.86h/g)
h = 0.43h
Now we solve for h:
1h = 0.43h
Subtracting 0.43h from both sides:
0.57h = 0
Divide both sides by 0.57:
h = 0
Therefore, the height of the fall is 0.
In conclusion:
(a) The time taken for the object to fall is √(0.86h/g), where h represents the height of the fall.
(b) The height of the fall is 0.