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>
Answer:
0.19m/s²
Explanation:
Initial velocity(u) = 50×1000/60×60
=13.88 m/s
Final velocity(v) = 36.5×1000/60×60
=10.13 m/s
Acceleration(a) = v-u/t
=10.13-13.88/19.5
a= -0.19m/s²
-a = 0.19m/s²
The magnitude of retar dation is 0.19m/s²
The outer planets<span> are further away, larger and made up mostly of gas. The </span>inner planets<span> (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars. After an asteroid belt comes the </span>outer planets<span>, Jupiter, Saturn, Uranus and Neptune.</span>
Answer:
a PDF is what u use to upload an assignment to turn it in to get graded
Answer:
Potential energy will be 
Explanation:
We have given the height of the basin is h = 6 m
Area of the basin 
Volume 
Density 
We know that mass is given by 
We know that potential energy is given by 
