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Suppose Ball 1 is at a height of 19.6 m when it reaches the top of its trajectory. At the instant Ball 1 is at the top of its trajectory, Ball 2 is released from rest at the same height. How much time does it take for either ball to reach the ground? Note that the vertical component y of the ball’s displacement points downward, in the -y direction.

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tochjosh

Answer:

Both ball will take approximately 2 sec to reach the ground.

Explanation:

Height reached by ball 1 at the top of its trajectory = 19.6 m

Assumptions:

  • As ball 1 gets to the top of its trajectory, its velocity becomes zero.
  • since not stated otherwise, we assume that both balls have the same properties
  • If ball 2 is released at this instance, then they both will travel down at the same time interval and have initial velocity as zero
  • since vertical travel is downwards, then acceleration due to gravity g is positive for both balls.

Using Newton's equation of motion,

s = ut + [tex]\frac{1}{2}[/tex]g[tex]t^{2}[/tex]

where s is the distance traveled down = 19.6 m

t is time taken to travel down = ?

g is acceleration due to gravity = 9.81 [tex]ms^{-2}[/tex]

u is the initial velocity which is zero for both balls

substituting, we have

19.6 = (0 x t) +( [tex]\frac{1}{2}[/tex] x 9.81 x [tex]t^{2}[/tex])

19.6 = 0 + 4.905[tex]t^{2}[/tex]

[tex]t^{2}[/tex] = 3.996

t = 1.99 sec ≅ 2 sec

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