Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

please show or justify all answers. work must be shown for full credit.…

Question

please show or justify all answers. work must be shown for full credit.

  1. a discus is thrown from a height of 4 feet with an initial velocity of 65 ft/s at an angle of 44° with the horizontal. how long will it take for the discus to reach the ground?

(show work)

  1. a rock is tossed from a height of 2 meters at an initial velocity of 30 m/s at an angle of 20° with the ground. write parametric equations to represent the path of the rock.

(show work)

  1. an airplane is taking off headed due north with an air speed of 173 miles per hour at an angle of 18° relative to the horizontal. the wind is blowing with a velocity of 42 miles per hour at an angle of s47°e. find a vector that represents the resultant velocity of the plane relative to the point of takeoff. let i point east, j point north, and k point up.

(show work)

  1. a 2600-pound truck is stopped at a red light on a hill with an incline of 25°. ignoring the force of friction, what force is required to keep the truck from rolling down the hill?

(show work)

Explanation:

Problem 4:

Step1: Identify vertical motion equation

The vertical motion of a projectile is given by \( y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2 \), where \( y_0 = 4 \) ft, \( v_{0y}=v_0\sin\theta = 65\sin(44^\circ) \) ft/s, and \( g = 32 \) ft/s² (acceleration due to gravity). We need to find \( t \) when \( y(t) = 0 \).

Step2: Substitute values into the equation

\( 0 = 4 + 65\sin(44^\circ)t - 16t^2 \)

First, calculate \( \sin(44^\circ) \approx 0.6947 \), so \( 65\sin(44^\circ) \approx 65\times0.6947 \approx 45.1555 \)

The equation becomes \( -16t^2 + 45.1555t + 4 = 0 \) or \( 16t^2 - 45.1555t - 4 = 0 \)

Step3: Solve quadratic equation

Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 16 \), \( b=-45.1555 \), \( c = -4 \)

Discriminant \( D = (-45.1555)^2 - 4\times16\times(-4) \approx 2039.02 + 256 = 2295.02 \)

\( \sqrt{D} \approx 47.91 \)

\( t=\frac{45.1555\pm47.91}{32} \)

We take the positive root: \( t=\frac{45.1555 + 47.91}{32} \approx \frac{93.0655}{32} \approx 2.91 \) seconds (the negative root is discarded as time can't be negative)

Step1: Find horizontal and vertical components of velocity

For a projectile, the horizontal component \( v_{0x}=v_0\cos\theta \), vertical component \( v_{0y}=v_0\sin\theta \), where \( v_0 = 30 \) m/s, \( \theta = 20^\circ \), \( y_0 = 2 \) m, \( g = 9.8 \) m/s².

Step2: Write parametric equations

Horizontal motion (constant velocity): \( x(t)=v_{0x}t = 30\cos(20^\circ)t \)

Vertical motion (with gravity): \( y(t)=y_0 + v_{0y}t - \frac{1}{2}gt^2 = 2 + 30\sin(20^\circ)t - 4.9t^2 \)

Calculate \( \cos(20^\circ)\approx0.9397 \), \( \sin(20^\circ)\approx0.3420 \)

So \( x(t)\approx28.19t \), \( y(t)=2 + 10.26t - 4.9t^2 \)

Step1: Decompose airplane's velocity

Airplane velocity: \( \vec{v}_{plane} \). Airspeed \( v = 173 \) mph, angle \( 18^\circ \) with horizontal, heading north.

Horizontal (north-east) components: \( v_{plane,x}=0 \) (no east component), \( v_{plane,y}=173\cos(18^\circ) \) (north), \( v_{plane,z}=173\sin(18^\circ) \) (up)

Step2: Decompose wind's velocity

Wind velocity: \( \vec{v}_{wind} \). Speed \( 42 \) mph, angle \( S47^\circ E \) (south of east by \( 47^\circ \)). So:

\( v_{wind,x}=42\cos(47^\circ) \) (east), \( v_{wind,y}=-42\sin(47^\circ) \) (south, negative north), \( v_{wind,z}=0 \) (no vertical component)

Step3: Find resultant velocity components

\( v_x = 0 + 42\cos(47^\circ) \), \( v_y = 173\cos(18^\circ) - 42\sin(47^\circ) \), \( v_z = 173\sin(18^\circ) + 0 \)

Calculate:

\( \cos(47^\circ)\approx0.6820 \), \( \sin(47^\circ)\approx0.7314 \), \( \cos(18^\circ)\approx0.9511 \), \( \sin(18^\circ)\approx0.3090 \)

\( v_x \approx 42\times0.6820 \approx 28.64 \)

\( v_y \approx 173\times0.9511 - 42\times0.7314 \approx 164.54 - 30.72 \approx 133.82 \)

\( v_z \approx 173\times0.3090 \approx 53.46 \)

Resultant velocity: \( \vec{v} = v_x\vec{i} + v_y\vec{j} + v_z\vec{k} \approx 28.64\vec{i} + 133.82\vec{j} + 53.46\vec{k} \)

Answer:

Approximately \( \boldsymbol{2.91} \) seconds (or more precise value depending on calculation)

Problem 5: