Afootball is thrown horizontally with an initial velocity of (16.6 m/s)x^. ignoring air resistance, the average acceleration of the football over any period of time is (-9.81 m/s2)y^.find the magnitude of the velocity at this time.

Solution:

According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.

So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.  

v_f=v_o + at ……..(a)

 [where v_f and v_o are final velocity and initial velocity, respectively]

Now ,

Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.  

Applying this value in equation (a)  

v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction  

For calculating the magnitude of the equation we have to square root the given value

         (16.6i - 17.165y)  

\\ \left | V  \right |=sqrt{16.6^{2}+17.165^{2}}\\ = \sqrt{275.56+294.637225}\\= \sqrt{570.197225}\\= 23.87[/tex]

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

we need to know the time interval during which the force acts. impulse of force = momentum change