a certain airplane has a speed of 80.5 m/s and is diving at an angle of 30.0° below the horizontal when the…

a certain airplane has a speed of 80.5 m/s and is diving at an angle of 30.0° below the horizontal when the pilot releases a radar decoy. the horizontal distance between the release point and the point where the decoy strikes the ground is d = 700 m.\nhow long is the decoy in the air?\nhow high was the release point?\nstep 1\ncriteria\nextraction of physical properties using a well - labelled diagram\nexcellent\nall information represented on a well labelled diagram including coordinate system\ndraw a labelled diagram for the problem (5 mins)\ntry to be systematic about your approach

a certain airplane has a speed of 80.5 m/s and is diving at an angle of 30.0° below the horizontal when the pilot releases a radar decoy. the horizontal distance between the release point and the point where the decoy strikes the ground is d = 700 m.\nhow long is the decoy in the air?\nhow high was the release point?\nstep 1\ncriteria\nextraction of physical properties using a well - labelled diagram\nexcellent\nall information represented on a well labelled diagram including coordinate system\ndraw a labelled diagram for the problem (5 mins)\ntry to be systematic about your approach

Answer

Explanation:

Step1: Analyze the horizontal - motion

The horizontal component of the initial velocity is $v_{0x}=v_0\cos\theta$, where $v_0 = 80.5$ m/s and $\theta = 30^{\circ}$. In horizontal - motion (with no acceleration, $a_x = 0$), the horizontal displacement is given by $x = v_{0x}t$. So, $t=\frac{x}{v_{0x}}$. First, calculate $v_{0x}=80.5\cos30^{\circ}=80.5\times\frac{\sqrt{3}}{2}\approx69.7$ m/s. Then, since $x = 700$ m, $t=\frac{700}{69.7}\approx10.0$ s.

Step2: Analyze the vertical - motion

The vertical component of the initial velocity is $v_{0y}=v_0\sin\theta=80.5\sin30^{\circ}=40.25$ m/s. The vertical displacement $y$ (height of the release - point) is given by the equation $y=v_{0y}t+\frac{1}{2}gt^{2}$, where $g = 9.8$ m/s² and $t = 10.0$ s. Substitute the values: $y=(40.25\times10)+\frac{1}{2}(9.8)(10)^{2}=402.5 + 490=892.5$ m.

Answer:

  1. The decoy is in the air for approximately $t = 10.0$ s.
  2. The height of the release - point is approximately $y = 892.5$ m.