A ball is dropped from a height x. It travels 122.5m in last second before hitting the ground. find x.

To find the height from which the ball was dropped, we can use the equations of motion under uniform acceleration due to gravity. When an object is dropped, it accelerates downwards at approximately 9.81 m/s². The distance traveled by the ball in the last second before it hits the ground is given as 122.5 m. Let's break this down step by step.

Understanding the Motion of the Ball

When an object falls freely, the distance it covers can be calculated using the formula:

• s = ut + (1/2)gt²

Where:

• s = distance traveled
• u = initial velocity (which is 0 for a dropped ball)
• g = acceleration due to gravity (approximately 9.81 m/s²)
• t = total time of fall

Distance in the Last Second

The distance traveled in the last second can also be expressed as:

• s_last = (1/2)g(t²) - (1/2)g((t-1)²)

Here, s_last is the distance traveled in the last second, and t is the total time of fall. We know that s_last = 122.5 m.

Setting Up the Equation

Substituting the known values into the equation:

• 122.5 = (1/2)(9.81)(t²) - (1/2)(9.81)((t-1)²)

This simplifies to:

• 122.5 = 4.905t² - 4.905(t² - 2t + 1)

Expanding the equation gives:

• 122.5 = 4.905t² - 4.905t² + 9.81t - 4.905

Which simplifies to:

• 122.5 = 9.81t - 4.905

Solving for Time

Now, we can solve for t:

• 122.5 + 4.905 = 9.81t
• 127.405 = 9.81t
• t = 127.405 / 9.81 ≈ 12.98 seconds

Finding the Height x

Now that we have the total time of fall, we can find the height from which the ball was dropped using the initial distance formula:

• x = (1/2)gt²

Substituting the values:

• x = (1/2)(9.81)(12.98)²

Calculating this gives:

• x = 0.5 * 9.81 * 168.48 ≈ 825.57 m

Final Result

Thus, the height from which the ball was dropped is approximately 825.57 meters.