Question

A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the tables edge in the

horizontal direction.
a. How long is the marble in the air?
b. What is the speed of the marble when it leaves the tables edge?
Im/s
C. What is its speed when it hits the floor?

Answer 1

The marble is in the air for approximately 0.5 seconds. It leaves the table's edge with a speed of 8 m/s and hits the floor with the same speed.

Step-by-step explanation:

To calculate the time the marble is in the air, we can use the formula t = √(2h/g), where h is the height of the table and g is the acceleration due to gravity. Plugging in the values, we get t = √(2*1.15/9.8) = 0.5 seconds.

To find the speed of the marble when it leaves the table's edge, we can use the equation v = d/t, where d is the horizontal distance and t is the time. Plugging in the values, we get v = 4/0.5 = 8 m/s.

When the marble hits the floor, its speed will be the same as its speed when it left the table's edge, which is 8 m/s.