The given problem can be represented using the following diagram:
We will use the non-parametric equation for the projectile motion:

Where:

Since the initial height is zero, we have:

Now, we substitute the known values:

Simplifying we get:

Now, we need to determine the value of "x" for which the value of "y" is 2.7 cm or in meters 0.027m. Substituting we get:

Now, we bring the terms on the right side to the left side by changing their signs:

Now, we divide both sides by 2.1:

we get an equation of the form:

the solution is given by the quadratic formula as follows:
![x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/rxvf73usjbbwyik14knxdemoz21vfz2ufc.png)
Substituting the values we get:
![x=\frac{-(-0.66)\pm\sqrt[]{(-0.66)^2-4(0.013)}}{2}](https://img.qammunity.org/2023/formulas/physics/college/2c5qdl2ksu2sha6i0tsr15sk1sh9lw8ial.png)
solving the operations inside the radical we get:
![x=\frac{-(-0.66)\pm\sqrt[]{(0.38)}}{2}](https://img.qammunity.org/2023/formulas/physics/college/mbxurnjb540otw5jbkww0q7eo94zz7y3e1.png)
Now, we solve for the positive value:
![x=\frac{-(-0.66)+\sqrt[]{(0.38)}}{2}](https://img.qammunity.org/2023/formulas/physics/college/8jz3ltr7fyk5rmf00g2a6i982xhyx58869.png)
Solving the operations:

Now, we solve for the negative value:
![x=\frac{-(-0.66)-\sqrt[]{(0.38)}}{2}](https://img.qammunity.org/2023/formulas/physics/college/kzmek3032zoghglsenqy6cgg3cscy4f3xa.png)
Solving the operations:

From the two possible values we take the smaller value, therefore, the horizontal reach is:

Or its equivalent 2 cm.
Part B. To determine the time we will use the following equation of motion for the projectile motion:

Now, we solve for "t":

Now, we substitute the values:

Solving the operations:

Therefore, the time is 0.013 seconds.