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A student drops a rock from rest at a distance h above the ground such that the rock hits the ground at a time t0 . At what distance above the ground should the rock be dropped such that it hits the ground at a time 2t0 after it is released.?

Respuesta :

onyebuchinnaji

The distance above the ground the rock should be raised in order to hits the ground double of initial time is 4h.

The given parameters;

  • initial velocity of the rock, u = 0
  • height above the ground, = h
  • time of motion of the rock from the height = t₀

The time of motion of the rock from the given height is calculated as;

[tex]h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt_0^2\\\\h = \frac{1}{2} gt_0^2\\\\t_0^2 = \frac{2h}{g} \\\\t_0 = \sqrt{\frac{2h}{g} }[/tex]

When the time of motion is doubled to 2t₀, the height above the ground should be;

[tex]2(t_0) = 2(\sqrt{\frac{2h}{g} } )\\\\2(t_0) = \sqrt{\frac{4\times 2h}{g} }[/tex]

Thus, the distance above the ground the rock should be raised in order to hits the ground double of initial time is 4h.

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