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Shkiper50 [21]
3 years ago
11

Fighter jet starting from airbase A flies 300 km east , then 350 km at 30° west of north and then 150km north to arrive finally

at airbase b
Physics
1 answer:
Tems11 [23]3 years ago
4 0
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North. 
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6 0
3 years ago
PLEASE HELP!!!!
defon

Answer:

v = 2.45 m/s

Explanation:

first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height of cliff = 15 m

Vi = Initial Vertical Velocity = 0 m/s

t = time taken = ?

g = 9.8 m/s²

Therefore,

15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²

t² = (15 m)/(4.9 m/s²)

t = √3.06 s²

t = 1.75 s

Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,

s = vt

where,

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v = original horizontal velocity = ?

Therefore,

4.3 m = v(1.75 s)

v = 4.3 m/1.75 s

<u>v = 2.45 m/s</u>

8 0
3 years ago
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