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a ball rolls off the edge of a table 1.44m above the floor and strikes the floor at a point 2m horizontally from the edge of the table. what is the time the ball was in the air?



Answer :

jcherry99

Answer:

t = 0 1.697 seconds

Explanation:

Comment

This is a standard question where the time is the same for both the vertical distance the ball has to drop and the time it takes to go horizontally.

Formulas

dv = vi*t + 1/2 a t^2

dv is the vertical distance that the ball has to travel in a downward direction

  • vi  is the initial vertical velocity. (which is zero.)
  • t is the time it takes to hit the floor
  • a is the gravitational acceleration

dh = vh * t

  • dh is the horizontal distance
  • vh is the horizontal velocity
  • t is the same time as the above formula

Givens

a = 9.81

dv = 1.44

vi = 0

t = ?

dh = 2 m

t = same as above (but not given)

Solution

dv = 1/2 a * t^2

1.44 = 1/2 * 9.81 *t^2                 Multiply by 2

2*1.44 = 1/2 * 2 * t^2

2.88 = t^2

t = 1.697 seconds

Note that you do not have to use the horizontal information to get the time. You only have to realize that the ball has no vertical velocity to begin with.

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