Answer:
d = 2021.6 km
Explanation:
We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them
Airplane 1
Height y₁ = 800m
Angle θ = 25°
cos 25 = x / r
sin 25 = z / r
x₁ = r cos 20
z₁ = r sin 25
x₁ = 18 103 cos 25 = 16,314 103 m
= 16314 m
z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m
2 plane
Height y₂ = 1100 m
Angle θ = 20°
x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m
z₂ = 20 103 without 25 = 8.452 103 m = 8452 m
The distance between the planes using the Pythagorean Theorem is
d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2
Let's calculate
d² = (18126-16314)² + (1100-800)² + (8452-7607)²
d² = 3,283 106 +9 104 + 7,140 105
d² = (328.3 + 9 + 71.40) 10⁴
d = √(408.7 10⁴)
d = 20,216 10² m
d = 2021.6 km
Brian’s Complexity Brian’s Complexity Brian’s Complexity Brian’s Complexity
Answer:
The distance the ball moves up the incline before reversing its direction is 3.2653 m.
The total time required for the ball to return to the child’s hand is 3.2654 s.
Explanation:
When the girl is moving up:
The final velocity (v) = 0 m/s
Initial velocity (u) = 4 m/s
a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).
Let time be t to reach the top.
Using
v = u + a×t
0 = 4 - 2.45*t
t = 1.6327 s
Since, this is the same time the ball will come back. So,
<u>Total time to go and come back = 2* 1.6327 = 3.2654 s
</u>
To find the distance, using:
v² = u² + 2×a×s
0² = 4² + 2×(-2.45)×s
s = 3.2653 m
<u>Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.</u>