Question

a marble rolls off horizontally from the edge of tabletop 1.50 m above the floor. it strikes the floor 2.0m from the base of the table. (a) how long it take marble to reach the floor? (b) what is it initial velocity?

Answer 1

Answer on Question 73798, Physics, Other

Question:

A marble rolls off horizontally from the edge of tabletop $1.50\,m$ above the floor. It strikes the floor $2.0\,m$ from the base of the table.

a) how long it takes marble to reach the floor?

b) what is it initial velocity?

Solution:

a) We can find the time that the marble takes to reach the floor from the kinematic equation:

$$y = v_{0y}t + \frac{1}{2}gt^2,$$

here, $y = 1.50\,m$ is the height of the tabletop, $v_{0y} = 0$ is the initial vertical velocity of the marble, $g = 9.8\,m/s^2$ is the acceleration due to gravity (we choose the downwards as the positive direction, therefore, the acceleration due to gravity will be positive).

Then, we get:

$$y = \frac{1}{2}gt^2,$$ $$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \cdot 1.50\,m}{9.8\, \frac{m}{s^2}}} = 0.55\,s.$$

b) We can find the initial velocity of the marble from the formula:

$$x = v_{0x}t,$$

here, $x = 2.0\,m$ is the distance from the base of the table, $v_{0x}$ is the initial horizontal velocity of the marble, $t$ is time that the marble takes to reach the floor.

Then, we get:

$$v_{0x} = \frac{x}{t} = \frac{2.0\,m}{0.55\,s} = 3.64\, \frac{m}{s}.$$

Answer:

a) $t = 0.55\,s$

b) $v_{0x} = 3.64\, \frac{m}{s}$

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