# Average acceleration of a falling ball on contact with floor

• bryansonex
In summary, a ball falling from a height of 10m and rebounding to 2.5m with a contact duration of 0.01 sec has an average acceleration of 2100 m/s^2. The mistake in the attempted solution was not correctly calculating the square roots.
bryansonex

## Homework Statement

A ball falls on the surface from 10m height and rebounds to 2.5m.If duration of contact with floor is 0.01 sec,then average acceleration during contact is-
g = 10m/s^2

## Homework Equations

$$\alpha = \frac{v_2 + v_1}{t}$$

## The Attempt at a Solution

Now here
$$v_{1} = \sqrt{2gh_{1}$$

$$v_{1} = \sqrt{50}$$

and

$$v_{2} = \sqrt{2gh_{2}$$

$$v_{2} = \sqrt{200}$$

$$\alpha = \frac{\sqrt{200+50}}{0.01}$$

After solving all of this,I get the answer the answer as 1581.14 m/s^2,but the answer in the book is 2100 m/s^2.Am i doing anything wrong over here?

You'll get the answer if you do the calculation more carefully.
An estimate, without calculator:
sqrt(50) approx. 7
sqrt(200) is 10sqrt(2)= 14 (approx.)
7+14=21
21/0.01 = 2100

Your mistake: sqrt(50)+sqrt(200) is NOT sqrt(50+200)

ohh...thanks nasu

## 1. What is average acceleration of a falling ball on contact with floor?

The average acceleration of a falling ball on contact with the floor is the rate at which the velocity of the ball changes over a given period of time as it falls and comes into contact with the floor. It is measured in units of meters per second squared (m/s²).

## 2. How is average acceleration of a falling ball on contact with floor calculated?

The average acceleration of a falling ball on contact with the floor can be calculated by dividing the change in velocity (final velocity minus initial velocity) by the change in time (final time minus initial time). This can be represented by the formula: a = (vf - vi) / (tf - ti), where a is the average acceleration, vf is the final velocity, vi is the initial velocity, tf is the final time, and ti is the initial time.

## 3. What factors can affect the average acceleration of a falling ball on contact with floor?

The average acceleration of a falling ball on contact with the floor can be affected by factors such as air resistance, mass and shape of the ball, and the surface material of the floor. Air resistance can slow down the ball's acceleration, while a heavier or more aerodynamic ball may have a greater acceleration. The type of surface the ball comes into contact with can also impact its acceleration due to factors such as friction.

## 4. Is the average acceleration of a falling ball on contact with floor constant?

No, the average acceleration of a falling ball on contact with the floor is not constant. It changes as the ball falls due to the effects of gravity and other external forces. Initially, the ball's acceleration will be equal to the acceleration due to gravity (9.8 m/s²), but as it nears the floor, other factors such as air resistance and friction will impact its acceleration.

## 5. What is the difference between average acceleration and instantaneous acceleration?

The average acceleration of a falling ball on contact with the floor is the average rate at which the ball's velocity changes over a given period of time. Instantaneous acceleration, on the other hand, is the acceleration of the ball at a specific moment in time. It can be thought of as the average acceleration over an infinitesimally small time interval. As the ball falls and nears the floor, its instantaneous acceleration will change constantly due to the effects of gravity and other external forces.

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