Alan

Rewrite as:

(204-203) + (202-201) + ... + (6-5) + (4-3) + (2-1) 

      1        +       1       + ... +    1    +    1    +   1

There are 102 terms so the sum is 102.

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Does this help?

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Baseball

A. Horizontal component of velocity is vx = v*cos(43°)

    Vertical component of velocity is vy = v*sin(43°)

Horizontal velocity is constant so time to hit ground is t = 188/vx

In this time vertical distance travelled is -0.9m so:

-0.9 = vy*t - (1/2)*g*t² where g = 9.8m/s²

so: -0.9 = vy*188/vx - (1/2)*9.8*(188/vx)²

-0.9 = 188*tan(43°) - (1/2)*9.8*(188/cos(43°))²/v²

v² = (1/2)*9.8*(188/cos(43°))²/(188*tan(43°) + 0.9)  m²/s²

$${\mathtt{v}} = {\sqrt{{\frac{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{188}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{43}}^\circ\right)}}}\right)}^{{\mathtt{2}}}\right)}{\left({\mathtt{188}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{43}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.9}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{42.865\: \!673\: \!803\: \!496\: \!16}}$$

v ≈ 43 m/s

Plane

A. The angle between south and southwest is 45°. 

The x-component (east) of velocity is -16*cos(45°):  vx = -16/√2 m/s

The y-component (north) of velocity is -(16*sin(45°) + 45):  vy = -(16/√2 + 45) m/s

B. Magnitude:  v_P/E = √(vx² + vy²)

C.  Direction:  θ = tan^-1(vy/vx)  But this will give a result in the first quadrant (i.e. an angle between 0° and 90°).  You will need to convert this to an angle in the third quadrant (i.e. between 180° and 270°).

Chris wrote: "I understood your method, Alan......but, am I misreading this?? You say "r" at the "time of interest" = 1m "

Well spotted Chris - you are right:  r = 2/3 m.  The r = 1m is there because I inadvertently set the whole worksheet to round all numbers to the nearest integer, instead of just setting the final result to be rounded to the nearest integer.

The correct value is used in the calculation though.

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A somewhat simpler way of tackling at this problem is as follows:

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My eyes aren't capable of making this out clearly enough (even with magnifier!). I suggest you upload a magnified image of just question 1 for starters.

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Here's my contribution:

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Logs don't have to have just integer bases Melody.  ln is log to base e, for example, and e is definitely not an integer!

A couple of other examples:

$${{log}}_{{\mathtt{4.5}}}{\left({\mathtt{6}}\right)} = {\mathtt{1.191\: \!268\: \!130\: \!927\: \!555\: \!2}}$$

$${{log}}_{{\mathtt{2.345}}}{\left({\mathtt{7}}\right)} = {\mathtt{2.283\: \!167\: \!287\: \!297\: \!550\: \!4}}$$

$${{log}}_{{\mathtt{3.45}}}{\left({\mathtt{1.2}}\right)} = {\mathtt{0.147\: \!226\: \!542\: \!852\: \!363\: \!6}}$$

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