John M. answered • 02/03/18

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- Acceleration a = -9.8 m/s
^{2} - Initial velocity vi = 23.0m/s
- Since the ball travels up and eventually comes back down to the same level, the vertical distance y traveled by the football is zero. So y = 0
- Find the vertical velocity of the football. The ball has both a horizontal and a vertical velocity. We're interested in the initial vertical velocity, i.e., the speed of the ball immediately after it is kicked. Call this v
_{iy} - v
_{iy}= 23.0m/s * sin 45 = 23 * .707 = 16.26 m/s - Now use the kinematic equation d = v
_{i}t + 0.5*(a*t^{2}) - The distance d is just the vertical distance y, which is zero. Substituting:
- 0 = 16.26*t + 0.5* (-9.8* t
^{2}) - Need to solve for t, which is the time the football is in the air
- 0 = t(16.26 - 4.9t)
- Divide both sides by t, and so 0 = 16.26 - 4.9t
- now solve for t.
- 16.26 = 4.9t
- 16.26/4.9 = t
- t = 3.32 seconds
- The football is in the air for
**3.32 seconds**before it hits the ground.