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Kipish [7]
2 years ago
12

A shot is fired at an angle of 60 degree horizontal with Kinetic energy E. If air resistance is ignored, the K.E at the top of t

rajectory is:
0, E/8, E/4, E/2
Physics
1 answer:
Lapatulllka [165]2 years ago
3 0
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 .
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A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
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See below

Explanation:

Vertical position = 45 +  20 sin (30) t  - 4.9 t^2

 when it hits ground this = 0

               0 = -4.9t^2 + 20 sin (30 ) t + 45

                0 = -4.9t^2 + 10 t +45 = 0     solve for t =4.22 sec

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     use this value of 't' in the equation to calculate max height = 50.1 m

      it has  4.22 - 1.02 to free fall = 3.2 seconds free fall

           v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

      it will <u>also</u> still have horizontal velocity =  20 cos 30 = 17.32 m/s

        total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30  * t  =  20 *  cos 30  * 4.22 = 73.1 m

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