sigh this is killing me here...
ok rockets launch at 250 m/s @ an angle of 75.0 degrees. There's a mountain 2500m away from launch point that is 1800m high. There's a boat 610 m from the island. So i'm suppose to find out how much it'll clear the moutain and close it'll come close to the boat
blah blah
so i find Vxi and Vyi
Vxi = 64.7 m/s
Vyi = 241 m/s
then I find time using formula
Vyf=Vyi+a delta t
solve for delta t and get 24.6s
try to find height use formula delta y = Vyi dleta T - .5g delta T^2 and I get 2960m
yada yada find distance traveled horizontally
d=rt
double time already found and got 49.2s
solve and got
3180m
so then I get three points
(0,0) (3180,2960) (3180,0)
I then use the lagrange method to find the equation of line
y = y1(x-x2)(x-x3)/[(x1-x2)(x1-x3)] +
y2(x-x1)(x-x3)/[(x2-x1)(x2-x3)]+
y3(x-x1)(x-x2)/[(x3-x1)(x3-x2)]
solve and I got -.001X^2 +3.72X
yada yada plug into graphic calculator and calc max and get vetex that is way off. The vertex should be the second point. i divided 3180 by two to get midpoint on horizontal axis and max height i already calculated... so (1590,2960) should be vertex but i don't get that
sigh I request help please tell me what i did wrong
From the information you've provided, it seems like you're trying to calculate the trajectory of a rocket launched at a certain velocity and angle. However, there are a few errors in your calculations.
First, let's break down the problem to make it easier to understand.
Given:
- Rocket launch velocity: 250 m/s
- Launch angle: 75.0 degrees
- Distance from launch point to the mountain: 2500 m
- Height of the mountain: 1800 m
- Distance from the island boat: 610 m
To find out how much the rocket will clear the mountain:
1. Calculate the initial x- and y-components of the velocity (Vxi and Vyi) using the given launch velocity and angle.
- Vxi = 250 * cos(75.0) = 64.7 m/s
- Vyi = 250 * sin(75.0) = 241 m/s
2. Calculate the time it takes for the rocket to reach its highest point using the y-component of velocity (Vyi) and the acceleration due to gravity (g = 9.8 m/s^2).
- Vyi = Vyi - g * t
- 0 = 241 - 9.8 * t
- t = 241 / 9.8 ≈ 24.59 s
3. Use the calculated time (t) to find the rocket's maximum height (h) using the y-displacement formula.
- Δy = Vyi * t - 0.5 * g * t^2
- h = 241 * 24.59 - 0.5 * 9.8 * (24.59)^2 ≈ 2957.7 m
So, according to these calculations, the rocket should reach a maximum height of approximately 2957.7 m.
Now, to find out how close the rocket will come to the boat, we need to calculate the horizontal distance traveled (d).
4. Calculate the time it takes for the rocket to reach the boat's position using the x-component of velocity (Vxi) and the horizontal distance (d).
- d = Vxi * t
Note: Here, t is the total time it takes for the rocket to reach the boat, which is twice the time calculated earlier (t = 2 * 24.59 s = 49.18 s).
- d = 64.7 * 49.18 ≈ 3184.75 m
Therefore, according to these calculations, the rocket should travel a horizontal distance of approximately 3184.75 m and will come close to the boat.
Now, let's address the issue with finding the vertex of the trajectory. Instead of using the Lagrange method and a quadratic equation, we can find the vertex directly using the known values.
The x-coordinate of the vertex can be found by taking half of the horizontal distance traveled:
- x_vertex = d / 2 = 1592.38 m
The y-coordinate of the vertex is the maximum height reached by the rocket:
- y_vertex = h ≈ 2957.7 m
Therefore, the vertex of the trajectory should be approximately (1592.38, 2957.7) and not (1590, 2960) as you mentioned.
It's possible that you made a calculation error or used incorrect values when trying to find the vertex using the Lagrange method. I would recommend double-checking your calculations and verifying the formulas used.