Help with free falling, find h with ball displacing .47h in 1 second before fall

In summary, the ball is dropped from a height of approximately 66.43 meters and falls for approximately 2.68 seconds before hitting the ground. This is determined by using the formula xf-xo=vot + 0.5gt^2 and setting two equations equal to each other to solve for t and then plugging t back into the equation to solve for h.
  • #1
mathphysicnb
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Homework Statement


A ball is dropped from a height h. The ball displaces 0.47h meters from the time it hits one second before it drops to the ground and when it hits the ground. Find h.


Homework Equations


xf-xo=vot + 0.5gt^2


The Attempt at a Solution


h = ?
t1(at 1 second before ball hits ground) = 1
s(at one second before till it hits ground) = 0.47h

So, this is what it visually looks like:

o (A) t=0, v=0, a=g, h=0
| |
| v
|
|
|
| (B) t=t1, a=g, h=0.53h
|
|
| (C) t=t2, a=g, h=H

So, from A to B, the ball falls 0.53h meters. From A to C, the ball falls H meters. And From B to C, the ball has fallen 0.47h meters in 1 second, which comes from the problem statement. So the ball at B has fallen t seconds. And when the ball reaches C, it has fallen t + 1 seconds.

I need to find h...

So, I use the formula above for A to B and again for A to C:

0.53h = 0.5gt^2 => h= (0.5g/0.53)t^2 (A to B)

h = 0.5g(t+1)^2 (A to C)

I have set both equations equal to h. Now I solve for t by setting said equations equal to each other:

(0.5g/0.53)t^2 = 0.5g(t +1)^2 =>

t^2/0.53 = (t+1)^2 =>

0 = (1-(1/0.53))t^2 + 2t + 1

Using quadratic formula, I get approximately 2.68 seconds.

I plug this back into one of the formulas above to get h:

h = 0.5g(t+1)^2 =~ 66.43 meters.

Is this correct?

I have to enter this online and I literally have one more attempt left before whatever answer I put gets submitted. Please help me and let me know if this is correct.

Thank you.
 
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  • #2
Your answer is correct.
 

Related to Help with free falling, find h with ball displacing .47h in 1 second before fall

1. How do you calculate the acceleration due to gravity using the given information?

To calculate the acceleration due to gravity, we can use the formula a = (2Δd)/(t^2), where a is the acceleration, Δd is the change in distance, and t is the time. In this case, Δd is equal to 0.47h and t is equal to 1 second. Plugging those values into the formula, we get a = (2*0.47h)/(1^2) = 0.94h. Therefore, the acceleration due to gravity is 0.94h.

2. What is the value of h in this scenario?

The value of h cannot be determined with the given information. We would need to know the initial velocity or the final velocity of the ball in order to calculate the value of h.

3. Can you explain the concept of free fall?

Free fall is the motion of an object that is only affected by the force of gravity. This means that the object is not being pushed or pulled by any other forces. In free fall, the object's velocity increases as it falls, while its acceleration remains constant at a value of 9.8 m/s^2 (assuming no air resistance).

4. How does the displacement of the ball change over time during free fall?

The displacement of the ball increases as time goes on during free fall. This is because the ball's velocity is constantly increasing due to the acceleration caused by gravity. The displacement can be calculated using the formula d = 0.5at^2, where d is the displacement, a is the acceleration, and t is the time.

5. Is there a maximum speed that the ball can reach during free fall?

Yes, there is a maximum speed that the ball can reach during free fall. This is called the terminal velocity, and it is the point at which the force of air resistance on the ball is equal and opposite to the force of gravity. At this point, the ball's acceleration will become zero and it will continue to fall at a constant speed.

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