A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball is fired horizontally with an initial speed of 700 m/s. Assuming air resistance can be neglected, approximately how long will the cannonball be in flight before it hits the ground?

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Answer:

  Cannonball will be in flight before it hits the ground for 2.02 seconds

Explanation:

  Initial height from ground = 20 meter.

  We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 [tex]m/s^2[/tex], we need to calculate time when s = 20 meter.

  Substituting

         [tex]20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds[/tex]

  So it will take 2.02 seconds to reach ground.

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