Respuesta :
Answer:
[tex]\theta = 33.5 degree[/tex]
Explanation:
As we know that net displacement is
[tex]d = 3.25 km[/tex]
altitude is given as
[tex]h = 2.15 km[/tex]
so its horizontal displacement is given as
[tex]x^2 + h^2 = d^2[/tex]
[tex]x^2 + 2.15^2 = 3.25^2[/tex]
[tex]x = 2.44 km[/tex]
now we have
[tex]v_x = 280 cos\theta[/tex]
[tex]v_y = 280 sin\theta[/tex]
now in x direction we have
[tex]2.44 \times 10^3 = (280 cos\theta) t[/tex]
in y direction we have
[tex]2.15 \times 10^3 = (280 sin\theta) t + 4.9 t^2[/tex]
from above equations we have
[tex]2.15 \times 10^3 = 2.44 \times 10^3 tan\theta + 4.9 (\frac{2.44 \times 10^3}{280 cos\theta})^2[/tex]
by solving above equation
[tex]\theta = 33.5 degree[/tex]
The angle at which the bomb strikes the target is 33.53°
Data;
- The height of the aircraft = 2.15km = 2150m
- The target of the bomb = 3.25km = 3250m
- The distance of the aircraft from the target = x
- Velocity or speed of aircraft = 280 m/s
Using Pythagoras's Theorem, we can calculate the distance of the aircraft from the target.
[tex](3250)^2 = x^2 + 2150^2\\x^2 = (3250)^2 - (2150)^2\\x = 2437m[/tex]
Kinematic Equation
Using kinematic equation,
[tex]y = y_1 + x tan \theta - \frac{gx^2}{2v^2 cos^2 \theta} \\y = (2150)(2437)tan\theta - \frac{(9.8)(2437)^2}{2(280)^2cos^2\theta} \\0 = (2150)+(2437)tan \theta - (371.2)sec^2 \theta\\0 = (2150)+(2437) tan \theta - (371.2) sec^2 \theta\\0 = 2150 + (2437) tan \theta - 381.2(1+ tan^2 \theta)\\-371.2 tan^2 \theta + 2437 tan \theta + 1778.8 = 0\\[/tex]
Since this quadratic equation we can solve this using quadratic formula
tan θ = -0.662 and 7.23
let's find tan inverse of this to find the value of θ
[tex]tan \theta = -0.662\\\theta = -33.53^0\\\theta = 33.53^0[/tex]
or
[tex]tan \theta = 7.23\\\theta = tan ^-^1 = 7.23\\\theta = 82.12^0[/tex]
The angle at which the bomb strikes the target is 33.53°
Learn more on kinematic equation here;
https://brainly.com/question/12351668
