Question

A football is thrown with an initial velocity of 22.26 m/s at an angle of 45o above the
horizontal. What is the velocity in ``y´´ at 3 seconds after the ball is thrown?

Answer 1

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]