Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass MMM and radius RRR about an axis perpendicular to th
1 answer:
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Answer:
A)
D) (Speed magnitude)
α = 22.49° (Speed direction above the horizontal)
Explanation:
Conceptual analysis:
To solve this problem we consider the following concepts:
1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.
The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):
Formula (1)
Formula (2)
Where:
Vo: Initial velocity in m/s
: Initial angle above the horizontal in grades
2) The formula to calculate its velocity at any vertical position(y) is as follows:
Formula (3)
Where:
: Final speed component in vertical direction in m/s
: Initial speed component in vertical direction in m/s
g: acceleration due to gravity in m/s2
3) The formulas to calculate the projectile velocity components at any time (t) are:
Because the movement is uniform in the x direction (constant speed)
Formula (4) Because the movement is uniformly accelerated in the y direction
Known information:
We know the following data:
= 34.5º above the horizontal
Development of the problem:
Initial speed components(Vox, Voy), (Formula (1), Formula (2)
A) Maximum height (h):
When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):
D) Speed of the projectile 1.50s after firing
We replace t=1.5 s in the formula(4)
(Speed magnitude)
α = 22.49° (Speed direction above the horizontal)