A block is at rest. The coefficients of static and kinetic friction are s = 0.81 and k = 0.69, respectively. The acceleration of
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Answer:
Explanation:
As we know that position of maximum on the screen in diffraction pattern is given as
here we know that
position of third maximum due to red color light and position of fifth maximum of unknown wavelength are same
so we have
so we have
Consider the projectile launched at initial velocity V at angle θ relative to the horizontal.
Neglect wind or aerodynamic resistance.
The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.
The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g
In order for D (horizontal distance) to be maximum,
That is,
Because
, therefore cos(2θ) = 0.
This is true when 2θ = π/2 => θ = π/4.
It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.
Neglect wind or aerodynamic resistance.
The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.
The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g
In order for D (horizontal distance) to be maximum,
That is,
Because
This is true when 2θ = π/2 => θ = π/4.
It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.