Answer:
Angle is 55.52°
and Initial Speed is v=26.48 m/s
Explanation:
Given data

Applying the kinematics equations for motion with uniform acceleration in x and y direction
So

Put the value of v₀ from equation (1) to equation (2)
So

Put that angle in equation (1) or equation (2) to find the initial velocity
So from equation (1)

Answer:
taking a shower brushing your teeth and washing your hands
Explanation:
The other nation will get mad at the other nation and they could start a war
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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.