The force of earth's gravitational field is always directed downwards (towards the center of the earth. When the ball is thrown up, it is going against the earth's gravitational field and so, the earth's gravitational force pulls it back down, accelerating it downwards.
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 .
<span>Reducing the distance between them. In theory, also increasing the mass; but you can't really change the mass of an object. However, you can compare the forces if you replace an object by a different object, which has a different mass.
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i hope this will work..
The solution would be like
this for this specific problem:
V^2 = 2AS = 2FS/M
V = sqrt(2FS/M) =
sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps
So the speed of the arrow as it leaves the bow
is 42.5 mps.
I am hoping that this answer has
satisfied your query and it will be able to help you in your endeavor, and if
you would like, feel free to ask another question.
