# Tutor profile: Greg T.

## Questions

### Subject: Physics (Newtonian Mechanics)

A cannon shoots a cannonball at a $$32^o$$ angle from the horizontal. The ball leaves the cannon at 100m/s. What is the highest point the cannonball will get to with respect to the cannon? How long will will the cannonball be in the air before it hits the ground? Assume acceleration due to gravity is $$-9.81m/s^2$$

The initial velocity of the cannonball must be split up into $$x$$ and $$y$$ directions, using 100m/s as the resultant velocity. To find the initial $$y$$ velocity, $$v_{yo}$$: $$v_{yo} = \sin{(32)}*100m/s = 53m/s $$ We know the $$y$$ velocity at the peak of the ball's path, $$v_{yf}$$, is zero. This is because at the peak of the ball's path (the highest point), it transitions to a negative $$y$$ velocity. To go from positive to negative, it must pass through a moment of zero $$y$$ velocity. Knowing the values of $$v_{yo}$$ and $$v_{yf}$$, we can use the following kinematic equation to solve for the $$y$$ distance travelled: $$v_f^2 = v_o^2 + 2ad $$, where $$a=acceleration$$ and $$d = distance$$ Subbing in the values we know: $$(0m/s)^2 = (53m/s)^2 + 2* (-9.81m/s^2)*d$$ Solving: $$d = 143.17m$$ Therefore, the cannonball will climb to a height of $$143.17m$$ above the cannon,. There are a few kinematic equations we can use to find the total time the cannonball spends in the air. Let's use this one: $$v_f = v_o + at$$ Subbing in the values we know: $$(0m/s) = (53m/s) + (-9.81m/s^2)*t$$ Solving: $$t = 5.4s$$ Keep in mind this value only accounts for the path taken from the cannon to the peak of the cannonball's path. In order to determine the total time spent in the air, we must multiply this value by 2 in order to account for the time the cannonball spends after it reaches the peak of its path until it hits the ground. Therefore: $$t_{total} = 5.4s*2=10.8s$$ Done!

### Subject: Basic Math

$$ 3 * 11 + (12/4)^3 = ? $$

Use BEDMAS, also known as PEMDAS. First, calculate what's in the brackets (or parentheses): $$ 3 * 11 + (3)^3 $$ Second, calculate the exponential term: $$ 3 * 11 + 27 $$ Third, calculate any terms which are being multiplied or divided: $$ 33 + 27 $$ Fourth, combine the terms to get your answer: $$ 33 + 27 = 60 $$ Done!

### Subject: Algebra

First, solve the system of equations for $$x$$, $$y$$ and $$z$$. Then, substitute these three values into the imaginary expression and rationalize the denominator. System of Equations: $$ 2x - y + 4z = -38 $$ (1) $$ -4x - 6y - z = -5 $$ (2) $$ x + 2y - 3z = 26 $$ (3) Imaginary Expression: $$ \frac{-3 - xi}{y - 2zi} $$

First, solve equation (1) for $$y$$: $$ y = 2x + 4z + 38 $$ Sub this value for $$y$$ into equation (2): $$ -4x - 6(2x + 4z + 38) - z = -5 $$ Simplify the previous equation to put $$x$$ in terms of $$z$$: $$ -4x - 12x - 24z - 228 - z = -5 $$ $$ -16x - 25z = 223 $$ $$ x = -223/16 - 25/16z $$ Second, solve equation (1) for $$x$$: $$ x = -19 + y/2 -2z $$ Sub this value for $$x$$ into equation (2): $$ -4(-19 + y/2 - 2z) - 6y - z = -5 $$ Simplify the previous equation to put $$y$$ in terms of $$z$$: $$ 76 - 2y + 8z - 6y - z = -5 $$ $$ -8y + 7z = -81 $$ $$ y = 81/8 + 7/8z $$ Sub the values for $$x$$ and $$y$$ (both in terms of $$z$$) into equation (3) and solve for $$z$$: $$ (-223/16 - 25/16z) + 2(81/8 + 7/8z) - 3z = 26 $$ $$ -25/16z + 14/8z - 3z = 26 + 223/16 - 162/8 $$ $$ -45/16z = 315/16 $$ $$ z = -7 $$ We already have equations for $$x$$ and $$y$$ in terms of $$z$$. Therefore, simply plug $$z$$ into these equations to solve for $$x$$ and $$y$$: $$ x = -223/16 - 25/16z $$ $$ x = -223/16 - 25/16(-7) $$ $$ x = -3 $$ $$ y = 81/8 + 7/8z $$ $$ y = 81/8 + 7/8(-7) $$ $$ y = 4 $$ These values can be checked by subbing them into the original equations: $$ 2(-3) - (4) + 4(-7) = -38 $$ (1) $$ -4(-3) - 6(4) - (-7) = -5 $$ (2) $$ (-3) + 2(4) - 3(-7) = 26 $$ (3) Looks good! Now for the second part of the question. The imaginary expression, after substitution, becomes: $$ \frac{-3 + 3i}{4 + 14i} $$ To rationalize the denominator, we multiply both the numerator and the denominator by a value which will get rid of the imaginary part on the bottom. That is, if the denominator is of the form $$a + bi$$, we multiply by $$a - bi$$. Therefore, we multiply by $$4 - 14i$$: Numerator: $$ (-3 + 3i)(4 - 14i) = -12 + 12i + 42i - 42i^2 = -12 + 54i + 42 = 30 + 54i $$ Denominator: $$ (4 + 14i)(4 - 14i) = 16 + 56i - 56i - 196i^2 = 16 + 196 = 212 $$ So the imaginary expression becomes: $$ \frac{30 + 54i}{212} $$ There is now no imaginary component to the denominator. Done!

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