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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°.
A mass weighing 32 pounds stretches a spring 2 feet.
(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.
(b) How many complete cycles will the mass have completed at the end of 4 seconds?
Answer:
Period = seconds
8 cycles
Explanation:
A mass weighing 32 pounds stretches a spring 2 feet;
it implies that the mass (m) =
m=
= 1 slug
Also from Hooke's Law
2 k = 32
k =
k = 16 lb/ft
Using the function:
(because of the initial position being above the equilibrium position)
( as a result of upward velocity)
NOW, we have:
However;
means
also implies that:
Hence,
Period can be calculated as follows:
=
= seconds
How many complete cycles will the mass have completed at the end of 4 seconds?
At the end of 4 seconds, we have:
cycles
