homework 4 begin date: 8/17/2025 11:00:00 pm due date: 9/16/2025 11:59:00 pm end date: 9/16/2025 11:59:00…

homework 4 begin date: 8/17/2025 11:00:00 pm due date: 9/16/2025 11:59:00 pm end date: 9/16/2025 11:59:00 pm\nproblem 14: (6% of assignment value)\na projectile is launched towards a hill that is d = 301 m away. the launch angle is θ = 45.5° above the horizontal with an initial speed of v₀ = 76.4 m/s. the hill can be approximated as a plane sloped at φ = 34.3°. neglect air resistance.\npart (a)\nwrite an equation for y as a function of x, d, and φ for the line that defines the slope of the hill.\ny = \ncos(α) cos(φ) cos(θ) ( ) 7 8 9 home\ncotan(φ) cotan(θ) sin(α) ↑↓ ^↓ 4 5 6 -\nsin(φ) sin(θ) tan(φ) / * 1 2 3 -\ntan(θ) d g + - 0. end\nm v₀ x √() backspace del clear\nsubmit hint feedback i give up!\n1 submission(s) remaining\nhints: 4% deduction per hint. hints remaining: 1\nfeedback: 5% deduction per feedback.\npart (b)\nwrite an equation for y as a function of x, g, v₀, and θ of the trajectory of the projectile.\npart (c)\nwhat is the x - coordinate, in meters, of the landing spot of the projectile?

homework 4 begin date: 8/17/2025 11:00:00 pm due date: 9/16/2025 11:59:00 pm end date: 9/16/2025 11:59:00 pm\nproblem 14: (6% of assignment value)\na projectile is launched towards a hill that is d = 301 m away. the launch angle is θ = 45.5° above the horizontal with an initial speed of v₀ = 76.4 m/s. the hill can be approximated as a plane sloped at φ = 34.3°. neglect air resistance.\npart (a)\nwrite an equation for y as a function of x, d, and φ for the line that defines the slope of the hill.\ny = \ncos(α) cos(φ) cos(θ) ( ) 7 8 9 home\ncotan(φ) cotan(θ) sin(α) ↑↓ ^↓ 4 5 6 -\nsin(φ) sin(θ) tan(φ) / * 1 2 3 -\ntan(θ) d g + - 0. end\nm v₀ x √() backspace del clear\nsubmit hint feedback i give up!\n1 submission(s) remaining\nhints: 4% deduction per hint. hints remaining: 1\nfeedback: 5% deduction per feedback.\npart (b)\nwrite an equation for y as a function of x, g, v₀, and θ of the trajectory of the projectile.\npart (c)\nwhat is the x - coordinate, in meters, of the landing spot of the projectile?

Answer

Explanation:

Step1: Find slope - intercept form for hill

The slope of the line representing the hill is $\tan(\phi)$. The line passes through the origin in the coordinate system where the launch - point is at the origin and the hill starts at $x = d$. Using the point - slope form $y - y_1=m(x - x_1)$, with $y_1 = 0$, $x_1=d$ and $m=\tan(\phi)$, we get $y=\tan(\phi)(x - d)$.

Step2: Find projectile trajectory equation

The initial horizontal velocity is $v_{0x}=v_0\cos(\theta)$ and the initial vertical velocity is $v_{0y}=v_0\sin(\theta)$. The equations of motion for the projectile are $x = v_{0x}t=v_0\cos(\theta)t$ (so $t=\frac{x}{v_0\cos(\theta)}$) and $y = v_{0y}t-\frac{1}{2}gt^{2}$. Substituting $t$ into the $y$ - equation: [ \begin{align*} y&=v_0\sin(\theta)\frac{x}{v_0\cos(\theta)}-\frac{1}{2}g\left(\frac{x}{v_0\cos(\theta)}\right)^{2}\ y&=x\tan(\theta)-\frac{gx^{2}}{2v_0^{2}\cos^{2}(\theta)} \end{align*} ]

Step3: Find the landing spot

Set the equation of the hill $y=\tan(\phi)(x - d)$ equal to the equation of the projectile $y=x\tan(\theta)-\frac{gx^{2}}{2v_0^{2}\cos^{2}(\theta)}$. [ \tan(\phi)(x - d)=x\tan(\theta)-\frac{gx^{2}}{2v_0^{2}\cos^{2}(\theta)} ] [ \tan(\phi)x-\tan(\phi)d=x\tan(\theta)-\frac{gx^{2}}{2v_0^{2}\cos^{2}(\theta)} ] [ \frac{gx^{2}}{2v_0^{2}\cos^{2}(\theta)}+(\tan(\phi)-\tan(\theta))x+\tan(\phi)d = 0 ] This is a quadratic equation $ax^{2}+bx + c = 0$ where $a=\frac{g}{2v_0^{2}\cos^{2}(\theta)}$, $b=\tan(\phi)-\tan(\theta)$ and $c = \tan(\phi)d$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, we take the positive root since $x>0$.

Answer:

Part (a): $y=\tan(\phi)(x - d)$ Part (b): $y=x\tan(\theta)-\frac{gx^{2}}{2v_0^{2}\cos^{2}(\theta)}$ Part (c): Solve the quadratic equation $\frac{gx^{2}}{2v_0^{2}\cos^{2}(\theta)}+(\tan(\phi)-\tan(\theta))x+\tan(\phi)d = 0$ for $x$ using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = \frac{g}{2v_0^{2}\cos^{2}(\theta)}$, $b=\tan(\phi)-\tan(\theta)$ and $c=\tan(\phi)d$ and take the positive root.