Question

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

Answer 1

Answer:

The ball was 1.01 seconds in the air

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:

$\displaystyle t=\frac{d}{v}$

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:

$\displaystyle t=\frac{3}{2.97}$

$t=1.01\ sec$

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:

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