Question

Project Throwing a Ball Where? Name Fror this project you will pretend you are throwing a ball on a planet (or moon) other than Earth. You will analyze the motion of your throw using calculus. To receive full and/or partial credit you must have complete and correct work. Make sure you are using calculus to justify your answers. Include units where possible. You may use a calculator to help with calculations, but you must show work and include the set up for what you are typing into the calculator. All decimal answers must accurate to 3 decimal places. Part 1: The Set Up. Choose a planet or moon: (Must be unique) Jupiter What is the coefficient of gravity for your location (m/s2):g=2,36 g Initial Height: You will be throwing a ball overhand: For our purposes we will assume you are releasing the ball at a height that is equal to your height. (I know this is not exactly what would happen, but this will make our set up easier.) What is your height? Convert to meters: 1.854 m Initial Velocity: Choose an initial velocity from the range below. 12 m/s to 40 m/s. (Note: 40 m/s is about 90mph ) Initial Velocity: 1sm/h Angle of your throw: Choose an angle of 30∘,45∘ or 60∘. Convert your angle to radians: θ=8.785 Parametric Equations to model your throw. x(t)=v0​(cosθ)ty(t)=−21​gt2+v0​(sinθ)t+h0​​

Answer 1

Throwing a ball on Jupiter using Calculus, amazing! (Highschool Math)

Sounds like a challenging and exciting project.

To answer your question, let me break down the information you provided:

• Planet/Moon: Jupiter
• Coefficient of gravity: g=2.36 m/s^2
• Initial height: h0 = 1.854 m (equal to your height)
• Initial velocity range: 12 m/s to 40 m/s
• Angle of throw: θ = 30°, 45°, or 60° (convert to radians)

Using the given information, we can model the motion of your throw using the parametric equations:

x(t) = v0(cos(θ))t

y(t) = -(1/2)gt^2 + v0(sin(θ))t + h0

Where:

• x(t) is the horizontal distance traveled by the ball at time t
• y(t) is the vertical height of the ball at time t
• v0 is the initial velocity of the ball
• θ is the angle of the throw (in radians)
• g is the coefficient of gravity
• h0 is the initial height of the ball (equal to your height)

To solve for the motion of the ball, you can plug in the given values for v0, θ, g, and h0, and then use calculus to find the time at which the ball reaches its maximum height and the time at which it hits the ground. You can also use calculus to find the maximum height and horizontal distance traveled by the ball.

Let's start by setting up the problem. You have chosen Jupiter as the planet, and the coefficient of gravity for Jupiter is g = 2.36 m/s^2. Your height is 1.854 meters, and you will be throwing the ball overhand. You have chosen an initial velocity of 1 m/s and an angle of 30 degrees.

Using these values, we can use the parametric equations to model the throw:

x(t) = v0(cosθ)t

y(t) = -(1/2)gt^2 + v0(sinθ)t + h0

Where:

x(t) is the horizontal distance traveled by the ball at time t

y(t) is the vertical height of the ball at time t

v0 is the initial velocity of the ball

θ is the angle of the throw (in radians)

g is the coefficient of gravity

h0 is the initial height of the ball (equal to your height)

To solve this problem, we need to find the time at which the ball reaches its maximum height and the time at which it hits the ground. We can do this by setting y(t) equal to zero and solving for t:

0 = -(1/2)gt^2 + v0(sinθ)t + h0

Using the quadratic formula, we get:

t = (v0(sinθ) ± sqrt((v0(sinθ))^2 - 4(-1/2)(g)(h0))) / 2(-1/2)(g)

Plugging in the given values, we get:

t = (1(sin(30°)) ± sqrt((1(sin(30°)))^2 - 4(-1/2)(2.36)(1.854))) / 2(-1/2)(2.36)

Simplifying this equation, we get:

t = (0.5 ± 1.546) / (-1.18)

t = -1.28 seconds or t = 0.78 seconds

Since we are only interested in positive time, the time at which the ball hits the ground is approximately 0.78 seconds.

To find the horizontal distance traveled by the ball, we can plug in the values of v0, θ, and t into the equation for x(t):

x(t) = v0(cosθ)t

Plugging in the given values, we get:

x(t) = 1(cos(30°))(0.78)

x(t) = 0.67 meters

Therefore, the horizontal distance traveled by the ball is approximately 0.67 meters.