Answer:
the answer is 1.4 seconds by solving it
1.4s
Let a₁ and a₂ be accelerations of a ball (upward) and rod (downward), respectively
49/5 mg
mg
Clearly, from the diagram,
Now, for the ball,
27 mg = may
...(ii)
and for the rod, mg - T= maz
On solving Eqs. (i), (ii) and (iii), we get
0m/s (upward)
a= m/s (downward)
So, acceleration of ball w.r.t. rod ==+= (3g/29)m/s²
Now, displacement of ball w.r.t. rod when it reaches
the upper end of rod is 1m. Using second equation of motion, sut+at
→t=58/30 1.4s (approx)
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Answer:
the answer is 1.4 seconds by solving it
1.4s
Let a₁ and a₂ be accelerations of a ball (upward) and rod (downward), respectively
49/5 mg
mg
Clearly, from the diagram,
Now, for the ball,
27 mg = may
...(ii)
and for the rod, mg - T= maz
On solving Eqs. (i), (ii) and (iii), we get
0m/s (upward)
a= m/s (downward)
So, acceleration of ball w.r.t. rod ==+= (3g/29)m/s²
Now, displacement of ball w.r.t. rod when it reaches
the upper end of rod is 1m. Using second equation of motion, sut+at
→t=58/30 1.4s (approx)