A ball is dropped from a height h without air resistance. If the ball falls a distance of h/2 in a time t, how much time is required to fall the remaining h/2?

Could you show steps because I don't understand how to work the problem. Thank you.

starting from rest ... d = 1/2 a t^2

... t = √(2d / a)

doubling the distance ... 2d = 1/2 a T^2
... T = √(4d / a) = t √2

2nd "half" fall time is ... t (√2 - 1)

Sure, I'd be happy to help with that!

To solve this problem, we can use the equations of motion. The first step is to find the time it takes for the ball to fall a distance of h/2.

We know that the distance fallen by an object in free fall in time t is given by the equation:
d = (1/2) * g * t^2

Since the ball falls a distance of h/2 in time t, we can substitute these values into the equation:
h/2 = (1/2) * g * t^2

Next, we can rearrange the equation to solve for t:
t = sqrt((h/2) / (1/2 * g))

Now, let's find the time required to fall the remaining h/2. We can use the same equation with the remaining distance d = h/2:
h/2 = (1/2) * g * t_remaining^2

Rearranging the equation, we can solve for t_remaining:
t_remaining = sqrt((h/2) / (1/2 * g))

Since we are solving for the remaining time, we already have the expression for t_remaining. Therefore, the time required to fall the remaining h/2 is also sqrt((h/2) / (1/2 * g)).

Please note that g is the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.

I hope this explanation was helpful! If you have any more questions, feel free to ask.

To solve this problem, we can use the equations of motion for free-falling objects. The key equation we will use is:

h = (1/2) * g * t^2

Where:
- h is the distance fallen
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken

Given that the ball falls a distance of h/2 in a time t, we can substitute these values into the equation:

h/2 = (1/2) * g * t^2

Let's solve for t:

1. Multiply both sides of the equation by 2 to eliminate the fraction:

h = g * t^2

2. Divide both sides by g to isolate t^2:

t^2 = h / g

3. Take the square root of both sides to solve for t:

t = sqrt(h / g)

This is the time required for the ball to fall the remaining distance of h/2.

Sure! To solve this problem, we can use the equations of motion for an object in free fall.

The key to solving this problem is to recognize that the time it takes for a ball to fall a certain distance is independent of the distance itself. In other words, the time it takes for the ball to fall the first half of the distance (h/2) is the same as the time it takes for it to fall the remaining half (h/2).

Let's break down the problem into steps:

Step 1: Recognize the given information.
- The ball falls a distance of h/2 in a time t.

Step 2: Understand the concept of free fall.
- In the absence of air resistance, any object in free fall accelerates due to gravity at a constant rate, denoted as "g" which is approximately 9.8 m/s².

Step 3: Determine the time it takes for the ball to fall the first half of the distance.
- We know that the distance fallen is h/2, so we can use the formula for displacement in free fall: s = ut + (1/2)gt², where s is the displacement, u is the initial velocity, t is the time, and g is the acceleration due to gravity.
- Since the ball is dropped (initial velocity u = 0), the formula simplifies to s = (1/2)gt².
- We substitute h/2 for s and solve for t: h/2 = (1/2)gt².
- Rearranging the equation, we have t² = (h/2) / (1/2)g.
- Simplifying that further, we get t² = (h/g).
- Taking the square root of both sides, we get t = √(h/g).

Step 4: Use the time obtained in Step 3 to find the time it takes for the ball to fall the remaining half of the distance.
- Since the time it takes to fall the first half is the same as the time it takes for the remaining half, the answer is t as well.
- Therefore, the time required to fall the remaining h/2 is also √(h/g).

To summarize, the time required to fall the remaining h/2 is equal to the time calculated in Step 3, which is √(h/g).