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 .
Answer:
Explanation:
An insulator. You can see ceramic insulators on telephone poles and power poles if you look carefully. If you live in a city, somewhere in that city is a power station. The insulators are huge. They have to be. The currents are very large in many cases.
Answer:
Speed at which it will reach the ground is given as

Total time for which it will remain in air is given as
t = 6.3 s
Explanation:
As we know that the object is projected upwards with speed


now when it will reach the ground then we have

so we have


so we have

Now speed of the object when it reaches the ground is given as


