+33661 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: vP/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°).
+33661 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: vP/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°).
+33661 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
