Explanation:
what is asked give brief explanation about question
 
        
             
        
        
        
If a football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees, then the vertical component of the initial velocity would be 12.65 m/s 
<h3>What is Velocity?</h3>
The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.
As given in the problem A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees
The horizontal component of the velocity is given by 
Vx = Vcosθ
The vertical component of the velocity is given by
Vy = Vsinθ
As we have to find the vertical component of the velocity given that speed of 16.71 m/s at an angle of 49.21 degrees from the ground 
Vy = 16.71 × sin49.21°
Vy = 12.65 m/s 
Thus, the vertical component of the velocity would be 12.65 m/s 
Learn more about Velocity from here
brainly.com/question/18084516
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Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.
        
             
        
        
        
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth,  R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
                                   w = 2π / T
                                   w = 2π / 24*3600
                                   w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
                                  v1 = R*w
                                  v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
                                  v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
                                  π/2  ........... s
                                  x     ............ 1/5 s
                                  x = π/2*5 = 18°     
- The radius of the earth R' at point where θ = 18° from the equator is:
                                 R' = R*cos(18)
                                 R' = (6.37 * 10 ^6)*cos(18)
                                 R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
                               v2 = R'*w
                               v2 = (6058230.0088)*(7.27 * 10^-5)
                               v2 = 440.433 m/s