Question

A ball rolls off a table and falls 0.75 m to the floor, landing with a speed of 4.1 m/s. What is the magnitude of the acceleration of the ball just before it strikes the ground? What was the initial speed of the ball?

Answer 1

2.4 m/s

Step-by-step explanation:

To find the magnitude of the acceleration of the ball just before it strikes the ground, you can use the following kinematic equation:

v^2 = u^2 + 2as

Where:

- v is the final velocity (4.1 m/s)

- u is the initial velocity (which you want to find)

- a is the acceleration

- s is the distance it fell (0.75 m)

First, let's find the acceleration:

v^2 = u^2 + 2as

(4.1 m/s)^2 = u^2 + 2 * a * 0.75 m

16.81 m^2/s^2 = u^2 + 1.5as

Now, we need to find the acceleration:

a = (v^2 - u^2) / (2s)

a = (16.81 m^2/s^2 - u^2) / (2 * 0.75 m)

Next, we can calculate the acceleration.

a = (16.81 m^2/s^2 - u^2) / 1.5 m

To find u, we need to rearrange the equation:

u^2 = 16.81 m^2/s^2 - 1.5as

Now, plug in the values and solve for u:

u^2 = 16.81 m^2/s^2 - (1.5 * a * 0.75 m)

Now, you can calculate u:

u^2 = 16.81 m^2/s^2 - 1.125as

u^2 = 16.81 m^2/s^2 - 1.125 * (a)

Now, you can calculate u:

u^2 = 16.81 m^2/s^2 - (1.125 * a)

Now, calculate u:

u^2 = 16.81 m^2/s^2 - 1.125 * a

u^2 = 16.81 m^2/s^2 - 1.125 * (9.8 m/s^2)

u^2 = 16.81 m^2/s^2 - 11.025 m^2/s^2

u^2 = 5.785 m^2/s^2

Taking the square root of both sides to find u:

u = √(5.785 m^2/s^2)

u ≈ 2.4 m/s (approximately)

So, the magnitude of the acceleration just before it strikes the ground is approximately 9.8 m/s^2 (due to gravity), and the initial speed of the ball was approximately 2.4 m/s.