A simple pendulum is made by tying a 4 kg stone to a string 4.8m long. The stone

Question

A simple pendulum is made by tying a 4 kg stone to a string 4.8m long. The stone is projected perpendicular to the string, away from the ground, with the string at an angle of 66 degrees with the vertical. It is observed to have a speed of 8 m/s when it passes its lowest point. What was he speed of the stone in m/s at the moment of release? A simple pendulum is made by tying a 4 kg stone to a string 4.8m long. The stone is projected perpendicular to the string, away from the ground, with the string at an angle of 66 degrees with the vertical. It is observed to have a speed of 8 m/s when it passes its lowest point. What was he speed of the stone in m/s at the moment of release? A simple pendulum is made by tying a 4 kg stone to a string 4.8m long. The stone is projected perpendicular to the string, away from the ground, with the string at an angle of 66 degrees with the vertical. It is observed to have a speed of 8 m/s when it passes its lowest point. What was he speed of the stone in m/s at the moment of release?

Explanation / Answer

vertical distance travelled, h = L*(1 - cos(66))

= 4.8*(1 - cos(66))

= 2.8477 m

let u is the initial speed and v is the final speed,

Apply, v^2 - u^2 = 2*g*h

v^2 - 2*g*h = u^2

==> u = sqrt(v^2 - 2*g*h)

= sqrt(8^2 - 2*9.8*2.8477)

= 2.86 m/s

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