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
Based on your description, it seems like you have made a few errors in your calculations. Let's go through the steps again and identify where the mistake might have occurred.
First, let's calculate the initial velocity components. Given that the rocket launches at 250 m/s at an angle of 75.0 degrees, we can use trigonometry to find its vertical and horizontal velocities:
Vx = V * cos(theta) = 250 * cos(75.0) ≈ 64.65 m/s
Vy = V * sin(theta) = 250 * sin(75.0) ≈ 241.05 m/s
Next, let's calculate the time of flight using the vertical velocity:
Vyf = Vyi + a * delta t
0 = 241.05 - 9.8 * delta t
Solving for delta t, we get delta t ≈ 24.64 s.
Now, let's find the rocket's maximum height using the formula delta y = Vyi * delta t - 0.5 * g * delta t^2:
delta y = 241.05 * 24.64 - 0.5 * 9.8 * (24.64)^2 ≈ 2956.58 m
Therefore, the rocket's maximum height is approximately 2956.58 m.
Moving on to the horizontal distance traveled, we can use the equation d = Vx * delta t:
d = 64.65 * 24.64 ≈ 1592.77 m
So, the rocket travels approximately 1592.77 m horizontally.
To determine the rocket's trajectory, we can plot the points (0, 0), (1592.77, 2956.58), and (1592.77, 0) on a coordinate system. It seems like you attempted to fit a quadratic equation to these points using the Lagrange method. However, there might have been an error in your calculations.
Instead, you can use the equation y = h - (h/l^2) * x^2 to approximate the rocket's trajectory, where h is the maximum height (2956.58 m) and l is half of the horizontal distance (1592.77 m/2 = 796.39 m). This equation assumes a parabolic trajectory.
So, the equation of the rocket's trajectory is y = 2956.58 - (2956.58/(796.39)^2) * x^2.
To find the closest point to the boat, given that it is 610 m from the island, you can substitute x = 610 into the equation and calculate the corresponding y-value.
I hope this helps you identify any errors and find the correct answers. If you need further assistance, please let me know!