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:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
If we know the value of d and h, we can solve the equation for v:
[tex]\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}[/tex]
Since the horizontal speed is constant, we can calculate the time the ball was in the air by using the equation:
[tex]\displaystyle t=\frac{d}{v}[/tex]
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:
[tex]\displaystyle v=3\cdot\sqrt{\frac {9.8}{2\cdot 5}}[/tex]
[tex]\displaystyle v=3\cdot\sqrt{\frac {9.8}{10}}[/tex]
Calculating:
[tex]v=2.97\ m/s[/tex]
Now calculate the time:
[tex]\displaystyle t=\frac{3}{2.97}[/tex]
[tex]t=1.01\ sec[/tex]
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:
[tex]\displaystyle t=\sqrt{\frac {2h}{g}}=\sqrt{\frac {10}{9.8}}=1.01\ sec[/tex]
