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Rudiy27
3 years ago
13

a cannonball is fired with a speed of 76 m/s from the top of a cliff. It strikes the plane below with a speed of 89 m/s. if we n

eglect air friction, how high is the cliff
Physics
1 answer:
galina1969 [7]3 years ago
8 0

Assuming the cannonball is fired horizontally, its horizontal velocity stays at a constant 76 m/s. At the point it hits the ground, it has a speed of 89 m/s, so if its vertical velocity at that moment is v_y, we have

89\dfrac{\rm m}{\rm s}=\sqrt{\left(76\dfrac{\rm m}{\rm s}\right)^2+{v_y}^2}\implies{v_y}^2\approx2145\dfrac{\mathrm m^2}{\mathrm s^2}

(taking the negative square root because we take the downward direction to be negative)

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respectively; a is the acceleration; and \Delta x is the change in position of a body. In the cannonball's case, it starts with 0 vertical velocity and is subject to a downward acceleration with magnitude g=9.80\frac{\rm m}{\mathrm s^2}. So we have

2145\dfrac{\mathrm m^2}{\mathrm s^2}-0=-2g\Delta y\implies\Delta y\approx-109.44\,\mathrm m

(which is negative because we take the cannonball's starting position at the top of the cliff to be the origin) so the cliff is about 109 m high.

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Explanation:

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       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

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     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

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Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

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