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You plan to throw stones by using a sling that you whirl over your head. You will whirl the stone in a horizontal circle with a radius of 1.50 m at a height of 2.50 m above the ground. If you want the stone to hit a target on the ground a distance of 15.9 m in front of you and 1.50 m to the side (note that this number is the same as the radius of the circle), what must be the magnitude of the centripetal acceleration of the stone just before you release it

Respuesta :

Answer:

[tex]330.635[/tex] meter per second square

Explanation:

Given

Height ,h [tex]= 2.50[/tex] m

Radius, r [tex]= 1.50[/tex] m

Distance travelled,  [tex]d = 15.9[/tex] m

Since, the stone is released and it falls towards the sky, its initial velocity is zero

As per equation

[tex]h = ut + \frac{1}{2} gt^2\\[/tex], we will evaluate "t"

Substituting the given values in above equation, we get

[tex]2.5 = 0 *t + 0.5 * 9.81 * t^2\\t^2 = \frac{2.5}{0.5*9.81}\\ t = \sqrt{0.5096} \\t = 0.714[/tex]

Velocity

[tex]= \frac{d}{t} \\= \frac{15.9}{0.714}\\= 22.27[/tex]

centripetal acceleration

[tex]= \frac{v^2}{r} \\= \frac{22.27^2}{1.5}\\ = 330.635\\[/tex]

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