#1**+10 **

**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} = √(vx^{2} + vy^{2})

**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°).

Alan
Oct 25, 2014

#1**+10 **

Best Answer

**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} = √(vx^{2} + vy^{2})

**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°).

Alan
Oct 25, 2014

#3**+5 **

**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*t^{2 }where g = 9.8m/s^{2}

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}/v^{2}

v^{2} = (1/2)*9.8*(188/cos(43°))^{2}/(188*tan(43°) + 0.9) m^{2}/s^{2}

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

v ≈ 43 m/s

Alan
Oct 25, 2014