Answer:
Explanation:
Given that,
The position of a particle is given by :
Let us assume we need to find its velocity.
We know that,
So, the velocity of the particle is 
.
I need help to I have a big test and I need help so bad
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
Force]/[force] = Newon/Newton = 1
Answer:
20m/second
Explanation:
The reason the answer is 20m/second is because to find the speed of the ball in this question you have to divide the distance over the time giving you the result of 20m/second
