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.\na. how long is the decoy in the air?\nb. how high was the release point?\n\nep 1\ncriteria\nextraction of physical properties using a well - labelled diagram\n\nexcellent\nall information represented on a well labelled diagram including coordinate system\n\ndraw a labelled diagram for the problem (5 mins)\ntry to be systematic about your approach
Answer
Explanation:
Step1: Find horizontal - velocity component
The initial velocity of the decoy $v_0 = 80.5$ m/s and the angle of dive $\theta=30.0^{\circ}$. The horizontal - velocity component is given by $v_{0x}=v_0\cos\theta$. So, $v_{0x}=80.5\cos30^{\circ}=80.5\times\frac{\sqrt{3}}{2}\approx69.7$ m/s.
Step2: Calculate time - of - flight
In horizontal motion (where there is no acceleration, $a_x = 0$), the horizontal displacement $x = d=700$ m. Using the formula $x = v_{0x}t$, we can solve for the time $t$. Rearranging the formula gives $t=\frac{x}{v_{0x}}$. Substituting $x = 700$ m and $v_{0x}\approx69.7$ m/s, we get $t=\frac{700}{69.7}\approx10.0$ s.
Step3: Find vertical - velocity component
The initial vertical - velocity component is $v_{0y}=v_0\sin\theta$. So, $v_{0y}=80.5\sin30^{\circ}=80.5\times\frac{1}{2}=40.25$ m/s.
Step4: Calculate height of release point
In vertical motion, the acceleration $a = g = 9.8$ m/s². Using the kinematic equation $y=v_{0y}t+\frac{1}{2}gt^{2}$. Substituting $v_{0y}=40.25$ m/s, $t = 10.0$ s, and $g = 9.8$ m/s², we have $y=(40.25\times10)+\frac{1}{2}\times9.8\times10^{2}=402.5 + 490=892.5$ m.
Answer:
a. The decoy is in the air for approximately $10.0$ s. b. The release point was approximately $892.5$ m high.