<span>Newton's Third Law of Action-Reaction is that for each and every action that happens, there is an equal and opposite reaction to it. In the scenario of a roller coaster, this is when you push down on the seat of the roller coaster as it flies along and the seat pushes back against you.</span>
Answer:
1) The maximum jump height is reached at A. 
2) The maximum center of mass height off of the ground is B. 
3) The time of flight is C. 
4) The distance of jump is B. 
Explanation:
First of all we need to decompose velocity in its rectangular components, so

1) We use,
, as we clear it for
and using the fact that
at max height, we obtain 
2) We can use the formula
for
, so

3) We can use the formula
, to find total time of fligth, so
, as it is a second-grade polynomial, we find that its positive root is
4) Finally, we use
, as it has an additional displacement of
due the leg extension we obtain,
, aprox 
Answer:
4. 10.0 m/s²
Explanation:
I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:


II) using these two equations after substitution v₀=0; v=30 and L=45:


Answer:

Explanation:
Given the following data;
Frequency = 4.0 x 10⁹ Hz
Planck's constant, h = 6.626 x 10-34 J·s.
To find the energy of the electromagnetic wave;
Mathematically, the energy of an electromagnetic wave is given by the formula;
E = hf
Where;
E is the energy possessed by a wave.
h represents Planck's constant.
f is the frequency of a wave.
Substituting the values into the formula, we have;


(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671