Question

A ping pong ball rolls off a 5.Om high table and lands on the floor, 3.0m

away from the table. How long is the ball in the air?

Answer 1

The ball was 1.01 seconds in the air

Step-by-step explanation:

Horizontal Motion

When an object is thrown horizontally with a speed v from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:

`\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}`

If we know the value of d and h, we can solve the equation for v:

`\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}`

Since the horizontal speed is constant, we can calculate the time the ball was in the air by using the equation:

`\displaystyle t=(d)/(v)`

The ping pong ball rolls from a height of h=5 m and lands on the floor a distance of d= 3 m away from the table. Calculate the speed:

`\displaystyle v=3\cdot\sqrt{\frac {9.8}{2\cdot 5}}`

`\displaystyle v=3\cdot\sqrt{\frac {9.8}{10}}`

Calculating:

`v=2.97\ m/s`

Now calculate the time:

`\displaystyle t=(3)/(2.97)`

`t=1.01\ sec`

The ball was 1.01 seconds in the air

Note: The time does not depend on the distance d at all. If we borrow the equation from free-fall motion, the time can be calculated with:

`\displaystyle t=\sqrt{\frac {2h}{g}}=\sqrt{\frac {10}{9.8}}=1.01\ sec`