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serg [7]
4 years ago
5

The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far befor

e firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to H.
Physics
1 answer:
Mekhanik [1.2K]4 years ago
6 0

Answer:

The new height the ball will reach = (1/4) of the initial height it reached.

Explanation:

The energy stored in any spring material is given as (1/2)kx²

This energy is converted to potential energy, mgH, of the ball at its maximum height.

If the initial height reached is H

And the initial compression of the spring = x

So, mgH = (1/2)kx²

H = kx²/2mg

The new compression, x₁ = x/2

New energy of loaded spring = (1/2)kx₁²

And the new potential energy = mgH₁

mgH₁ = (1/2)kx₁²

But x₁ = x/2

mgH₁ = (1/2)k(x/2)² = kx²/8

H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)

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with v_0 and s_0 being the initial velocity and distance, both 0 in this case, and with "a" denoting the acceleration, in this case  solely due to gravitational acceleration so: "g."

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7 0
4 years ago
A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
Ray Of Light [21]

Answer:

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

  max height is at  t= - b/2a = 10/9.8 =1.02

     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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2 years ago
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4 years ago
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3 years ago
A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 120 meters. What is the approxi
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The approximate acceleration of the car would be <span>3.00

</span>How? We have to use the formula to find the velocity.

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