Jon watches his sister Jessica skateboard. When she jumps and launches herself into the air with an initial speed of v0 feet per second, her path in terms of time, t, in seconds, is represented by these equations.
Equation 1 xt v0tcos(theta)
Equation 2 yt v0tsin(theta) 16 t 2
The first equation models the horizontal distance, xt , that the skateboarder travels, and the second equation models the vertical height, yt , that the skateboarder attains.
1)Jessica attains a height of 4.7 feet above the launch and landing ramps after 1 second. Her initial velocity is 25 feet per second. To find the angle of her launch, which equation can you use with the given information to solve for θ? (Answer: 1 or 2).
A) arc sin (21.5/30)
B) arc cos (10.5/30)
C) arc cos (21.5/30)
D) arc sin (10.5/30)
+23246 You have two equations:
Equation #1: x(t) = v0·cos(θ) gives the x-value, which is a horizontal value
Equation #2: y(t) =v0·sin(θ) - 16t² gives the y-value, which is a vertical value
Question 1) says that the height is 4.7; this is a y-value, specifically y(1).
It also gives you the time, t = 1, and gives you the initial velocity, v0 = 25.
You now have the equation: y(1) = 4.7 = 25sin(θ) - 16(1)²
Question 2) Solve the equation from question 1, for θ.
Question 3) You are again looking for height: use y(t) =v0·sin(θ) - 16t²; entering, .5 for t, 25 for v0, and your answer for question 2 for θ. Solve for y(.5).
Question 4) I'm assuming, possibly incorrectly, that they want the x-value after 1 second. Use the first equation, substituting the values for v0 and θ.
Question 5) Redo question 2, but using 5.2 for y(t), 1 for t, and 28 for v0.
Question 6) Redo question 3, but using 28 for vo, .5 for t, and you answer for question 5 for θ.
Question 7) Redo question 4, for the numbers for Jon.
Question 8) Redo question 3 using the new number.
If you need more help, repost where your questions are.
+23246 You have two equations:
Equation #1: x(t) = v0·cos(θ) gives the x-value, which is a horizontal value
Equation #2: y(t) =v0·sin(θ) - 16t² gives the y-value, which is a vertical value
Question 1) says that the height is 4.7; this is a y-value, specifically y(1).
It also gives you the time, t = 1, and gives you the initial velocity, v0 = 25.
You now have the equation: y(1) = 4.7 = 25sin(θ) - 16(1)²
Question 2) Solve the equation from question 1, for θ.
Question 3) You are again looking for height: use y(t) =v0·sin(θ) - 16t²; entering, .5 for t, 25 for v0, and your answer for question 2 for θ. Solve for y(.5).
Question 4) I'm assuming, possibly incorrectly, that they want the x-value after 1 second. Use the first equation, substituting the values for v0 and θ.
Question 5) Redo question 2, but using 5.2 for y(t), 1 for t, and 28 for v0.
Question 6) Redo question 3, but using 28 for vo, .5 for t, and you answer for question 5 for θ.
Question 7) Redo question 4, for the numbers for Jon.
Question 8) Redo question 3 using the new number.
If you need more help, repost where your questions are.
