Max. Height reached by each ball (h) = 5 m.
Final velocity, v =0
Let Initial velocity=u m/s
=> v^2 = u^2 - 2gh
=> 0=u^2 - 2×10×5
=> u^2 = 100
=> u = 10 m/s
Now, put u=10 in first equation of motion to get the value of time,
V = u - gt
=>0 = 10 -10t
=> t= 1 sec.
Time taken by each ball to reach the max. height = 1 sec.
Conversely,
No. of balls thrown in 1 sec. = 1
No. of balls thrown in 1 min. or 60 sec. =60 balls.
Three identical balls are placed on a frictionless
horizontal surface touching each other. They stick to
each other because of adhesive. Another ball of same
radius and mass m is placed over the void created by
the three balls. Find the forces applied by the balls
kept on the floor to each other if the system remains
in equilibrium.
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A car loaded with water having total mass 1000 kg moves on a straight horizontal Road starting from rest under the action of a force of 100 Newton. The water spills throw a small hole in the bottom at a rate of 0.1 KG per second the velocity of cart after 300 seconds is nearly.
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