Question

If the ball bounces straight up and returns to the floor 2.2 s after first striking it, what was the ball's greatest height above the floor?

Answer 1

When a ball is thrown vertically upward and then returns to the ground, its motion can be described using the equations of motion. The formula to calculate the maximum height reached by the ball is given by:

`\[ \text{Max Height} = (1)/(2) g t^2 \]`

Where:

• ( g ) is the acceleration due to gravity (approximately 9.8 m/s²)
• ( t ) is the time taken for the ball to reach its maximum height (2.2 seconds in this case)

Plugging in the values:

`\sf{\[ \text{Max Height} = (1)/(2) * 9.8 \, \text{m/s}^2 * (2.2 \, \text{s})^2 \]}`

`= > \: \: \[ (2.2 \, \text{s})^2 = 4.84 \, \text{s}^2 \]`

Now plug this value into the equation and solve for the maximum height:

`\sf{\[ \text{Max Height} = (1)/(2) * 9.8 \, \text{m/s}^2 * 4.84 \, \text{s}^2 \]</p><p>\[ \text{Max Height} }\\ = 23.816 \, \text{m} \]`

So, the ball's greatest height above the floor is approximately 23.816 meters.