Question

A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much time does it take the marble to hit the ground? b. What was the velocity of the marble on the table?

Answer 1

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

$y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]$

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

$(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]$