An airplane is flying with a velocity of 220 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.8 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle θ?
I found
Vox=190.53
Voy=110
T=110/9.8=11.22
X=(190.53)(22.44)=4277.4
y=(110)(11.22)+.5(9.8)(11.22)^2
y=617.339
4277.2^2+617.339^2=c^2
C=4321.72
inv_sin(617.339/4321.72)=8.21
But my answer wasn't correct, What am I doing wrong?
hmm..
Vx= 220cos30 = 190.5
Vy = 110
s=-Vy*t + (1/2)*a*t^2
2800 = -110*t + (1/2)*(9.81)*t^2
use quadratic equation to solve for t
t=37.6s
v=g*t
v=9.81 * 37.6
v=368.85
tan^-1(v/vx)
tan^-1(368.85/110) = 73.39
See if that's correct..
oops
tan^-1(368.85/190.5) = 62.68
To find the angle θ, you need to use the horizontal and vertical displacement values of the flare when it hits the target on the ground.
You have correctly calculated the horizontal displacement (X) as 4277.4 m. However, you have made an error in calculating the vertical displacement (y). The correct equation for calculating the vertical displacement is:
y = Voy * t + 0.5 * g * t^2
where Voy is the initial vertical velocity of the flare (110 m/s), t is the time taken for the flare to hit the ground (which you have correctly found as 11.22 s), and g is the acceleration due to gravity (approximated as 9.8 m/s^2).
So, the correct calculation for y is:
y = (110 * 11.22) + 0.5 * (9.8) * (11.22)^2
= 1234.86
Now, to calculate the angle θ, you need to find the ratio of the vertical displacement (y) to the horizontal displacement (X), and then take the inverse tangent of that ratio:
θ = arctan(y / X)
= arctan(1234.86 / 4277.4)
≈ 16.89°
Therefore, the correct angle θ is approximately 16.89°.
To find the angle θ, we need to use the concept of projectile motion. Let's go through the steps again to identify any errors.
1. First, let's break down the velocity of the plane into horizontal and vertical components.
Vox (horizontal component) = acceleration * cosθ
Voy (vertical component) = acceleration * sinθ
Given that the velocity of the plane is 220 m/s at an angle of 30.0°, we can calculate the horizontal and vertical components:
Vox = 220 * cos(30) ≈ 190.53 m/s
Voy = 220 * sin(30) ≈ 110 m/s
2. Next, let's calculate the time of flight (T). We can use the formula:
T = 2 * Voy / acceleration
Given that the acceleration due to gravity is 9.8 m/s², we can calculate T:
T = (2 * 110) / 9.8 ≈ 22.45 s
3. Now, let's find the horizontal distance traveled by the flare (X) using the formula:
X = Vox * T
X = 190.53 * 22.45 ≈ 4277.78 m
4. Finally, let's find the vertical distance traveled by the flare (Y) using the formula:
Y = Voy * T - 0.5 * acceleration * T²
Y = (110 * 22.45) - (0.5 * 9.8 * 22.45²) ≈ 618 m
5. To find the angle θ, we can use the Pythagorean theorem to find the magnitude of the total displacement of the flare (C):
C = √(X² + Y²)
C = √(4277.78² + 618²) ≈ 4321.3 m
6. Finally, we can find the angle θ using the inverse sine function:
θ = arcsin(Y / C)
θ = arcsin(618 / 4321.3) ≈ 0.1437 radians
Converting to degrees:
θ ≈ 0.1437 * (180/π) ≈ 8.23°
So, the angle θ is approximately 8.23°.
Make sure to double-check your calculations and use the correct values for the trigonometric functions and the constant of acceleration due to gravity.