It is B: ionosphere. This has the ability to bounce radio signals
Answer:
<em>1.43 s.</em>
Explanation:
Using one of the equations of motion,
S = ut + 1/2gt².......................... Equation 1
Where S = height of the cliff, u = initial velocity, t = time, g = acceleration due to gravity.
<em>Note: When the rock begins to fall from the maximum height, u = 0 m/s, g = positive</em>
<em>Given: S = 10 m, u = 0 m/s</em>
<em>Constant: g = 9.8 m/s²</em>
<em>Substituting these values into equation,</em>
<em>10 = 0(t) + 1/2(9.8)(t²)</em>
<em>10 = 0 + 4.9t²</em>
<em>t² = 10/4.9</em>
<em>t² = 100/49</em>
<em>t = √(100/49)</em>
<em>t = 10/7</em>
<em>t = 1.43 s.</em>
<em>Thus the rock spend 1.43 s in air</em>
Initial velocity of a car is 36 km/h . Find the distance after min, if it goes with acceleration 2 m/s².
Initial velocity, u = 36 km/h
Time, t = 1min
Acceleration, a = 2m/s²
Apply 2nd equation of motion
Answer:
Time of flight=3.5 seconds
Speed at maximum height is 0
Explanation:
Φ=60°
initial velocity=u=20m/s
Acceleration due to gravity=g=9.8 m/s^2
Total time of flight=T
Final speed=v
question 1:
T=(2 x u x sinΦ)/g
T=(2 x 20 x sin60)/9.8
T=(2 x 20 x 0.8660)/9.8
T=34.64/9.8
T=3.5 seconds
Question 2
Speed at maximum height is 0
Answer: 
Explanation:
Given
final velocity at takeoff 
Acceleration of the plane can be 
Initial velocity is zero for the plane i.e. 
Using the equation of motion
![\Rightarrow v^2-u^2=2as\quad [\text{s=displacement}]\\\text{Insert the values}\\\Rightarrow (28.1)^2-0=2\times 2\times s\\\\\Rightarrow s=\dfrac{789.61}{4}\\\\\Rightarrow s=197.40\ m](https://tex.z-dn.net/?f=%5CRightarrow%20v%5E2-u%5E2%3D2as%5Cquad%20%5B%5Ctext%7Bs%3Ddisplacement%7D%5D%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5CRightarrow%20%2828.1%29%5E2-0%3D2%5Ctimes%202%5Ctimes%20s%5C%5C%5C%5C%5CRightarrow%20s%3D%5Cdfrac%7B789.61%7D%7B4%7D%5C%5C%5C%5C%5CRightarrow%20s%3D197.40%5C%20m)
Thu,s the minimum length must be
