Answer:
Explanation:
Given
mass of boy=36 kg
length of swing=3.5 m
Let T be the tension in the swing
At top point 
where v=velocity needed to complete circular path
Th-resold velocity is given by 
So apparent weight of boy will be zero at top when it travels with a velocity of 
To get the velocity at bottom conserve energy at Top and bottom
At top 
Energy at Bottom 
Comparing two as energy is conserved
Apparent weight at bottom is given by
Rubber tape is used to round sharp edges
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
