WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of
1 answer:
Answer:
82.1 km
Explanation:
We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:
Vector B is 48 km south, so:
Finally, vector C:
Now we add the components along each direction:
So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is
You might be interested in
Answer:
θ = tan⁻¹ ()
Explanation:
This problem must be solved using projectile launch ratios. Let's analyze the situation, the projectile explodes at the highest point, therefore we fear the height (i = h), the speed at this point is the same, but the direction changes, we are asked to find the smallest angle of the speed in the point of arrival with respect to the x-axis.
The speed at the arrival point (y = 0)
v² = vₓ² + v_y²
Let's see how this angle changes, for two extreme values:
* The particle that falls from the point of explosion, in this case the speed is vertical
v = v_y
the angle with the horizontal is 90º
* The particle leaves horizontally from the point of the explosion, the initial velocity is horizontal
vₓ = v
the final velocity for y = 0
v_f = vₓ² + v_y²
therefore the angle has a value greater than zero and less than 90º
As they ask for the smallest angle, we can see that we must solve the last case
the output velocity is horizontal vₓ = v
Let's find the velocity when it hits the ground y = 0, with y₀ = h
=
- 2 g (y-y₀)
v_{y}^2 = - 2g (0- y₀)
let's calculate
v_{y}^2 = 2 9.8 h
we use trigonometry to find the angle
tan θ =
θ = tan⁻¹ ()
let's calculate
θ = tan⁻¹ ()
Answer
given,
height of window = 4 m
time taken to travel = 1 s
acceleration due to gravity = 9.8 m/s²
u = -0.905 m/s
initial velocity of ledge v = 0
now,
v² = u² + 2 a s
(-0.905)² = 0 + 2 × 9.8 ×s
s = 0.042 m