A projectile travels $(100+2 a) \mathrm{m}$ in a direction of $(2 b+45)$ degrees from the horizontal. The projectile then travels a further $(50+2 c) \mathrm{m}$ in a direction of $(2 d+15)$ degrees from the horizontal (hint: draw a diagram to depict the motion of this projectile in the form of a triangle): $\newline$ (a) Find the angle to the horizontal of the final position of the projectile relative to its starting position; $\newline$ (i) using trigonometric rules [ $4$ Marks] $\newline$ (ii) using vector components [ $8$ Marks]

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Q. A projectile travels

(100+2 a) \mathrm{m}

in a direction of

(2 b+45)

degrees from the horizontal. The projectile then travels a further

(50+2 c) \mathrm{m}

in a direction of

(2 d+15)

degrees from the horizontal (hint: draw a diagram to depict the motion of this projectile in the form of a triangle):

\newline

(a) Find the angle to the horizontal of the final position of the projectile relative to its starting position;

\newline

(i) using trigonometric rules [

4

Marks]

\newline

(ii) using vector components [

8

Marks]

Understand Projectile Motion: To solve this problem, we first need to understand the motion of the projectile. The projectile makes two separate movements: the first one at an angle of $2b+45$ degrees and the second one at an angle of $2d+15$ degrees from the horizontal. We will draw a diagram to visualize the two vectors representing these movements.
Find Components Using Trigonometry: Using trigonometric rules, we can find the horizontal and vertical components of each movement. For the first movement, the horizontal component is $(100+2a)\cos(2b+45)$ and the vertical component is $(100+2a)\sin(2b+45)$ . For the second movement, the horizontal component is $(50+2c)\cos(2d+15)$ and the vertical component is $(50+2c)\sin(2d+15)$ .
Calculate Resultant Vector Components: We add the horizontal components and the vertical components separately to find the resultant vector's components. The resultant horizontal component $R_h$ is $(100+2a)\cos(2b+45) + (50+2c)\cos(2d+15)$ , and the resultant vertical component $R_v$ is $(100+2a)\sin(2b+45) + (50+2c)\sin(2d+15)$ .
Calculate Angle to Horizontal: The angle $\theta$ to the horizontal of the final position of the projectile relative to its starting position can be found using the arctangent of the vertical component over the horizontal component, $\theta = \arctan(\frac{R_v}{R_h})$ .
Calculate $\theta$ Using Given Values: Now we will calculate the angle $\theta$ using the given values. However, since we do not have specific values for $a$ , $b$ , $c$ , and $d$ , we cannot compute a numerical answer. Instead, we will leave the answer in terms of $a$ , $b$ , $c$ , and $d$ .
Represent Movements as Vectors: Using vector components, we can represent the first movement as a vector $V_1$ with components [(100+2a)\cos(2b+45), (100+2a)\sin(2b+45)\] and the second movement as a vector $V_2$ with components [(50+2c)\cos(2d+15), (50+2c)\sin(2d+15)\].
Calculate Magnitude of Resultant Vector: The resultant vector $R$ is the sum of vectors $V_1$ and $V_2$ . So, $R = [V_{1_x} + V_{2_x}, V_{1_y} + V_{2_y}]$ , where $V_{1_x}$ and $V_{2_x}$ are the horizontal components, and $V_{1_y}$ and $V_{2_y}$ are the vertical components of $V_1$ and $V_2$ , respectively.
Find Direction of Resultant Vector: The magnitude of the resultant vector $R$ can be found using the Pythagorean theorem: $|R| = \sqrt{R_h^2 + R_v^2}$ , where $R_h$ is the sum of the horizontal components and $R_v$ is the sum of the vertical components.
Find Direction of Resultant Vector: The magnitude of the resultant vector $R$ can be found using the Pythagorean theorem: $|R| = \sqrt{R_h^2 + R_v^2}$ , where $R_h$ is the sum of the horizontal components and $R_v$ is the sum of the vertical components.The direction of the resultant vector $R$ relative to the horizontal can be found using the arctangent function: $\theta = \arctan(\frac{R_v}{R_h})$ , where $R_v$ is the vertical component and $R_h$ is the horizontal component of the resultant vector $R$ .