Question

an arecanut took one second to reach the ground from the areca nut tree.then what will be the approximate height of the arecanut tree

Answer 1

The height of the Areca nut tree is 5 metre.

Explanation:

1. Here Areca nut is fall from tree. So at time of fall its velocity is zero.

2. This case is free fall, means there are only gravitational force acting on nut.

3. So acceleration of nut is gravitational acceleration. where value of gravitational acceleration(g) is 10$(\frac{m}{s^{2}})$.

4. Now from equation of motion for case of free fall.

  $H= u\times t+(\frac{1}{2})\times g\times t^{2}$    ....1)

  where

  H = height from which object is fall.

  u = initial velocity of object = 0$(\frac{m}{s})$

  t = time taken by object to cover the height = 1 (second)

  g = gravitational acceleration= 10$(\frac{m}{s^{2}})$

5. Now from equation 1)

   $H= 0\times 1+(\frac{1}{2})\times 10\times 1^{2}$

  On calculating

  H = 5 metre

   

Answer 2

From question, we have;

• Time taken by arecanut to reach the ground(t) = 1 sec

Now,

As, this thing is happening under the influence of gravity,thus the acceleration it will go through will be equal to accleration due to gravity.

• Acceleration due to gravity(a) = 9.8 m/s or 10 m/s(approx)

And, the arecanut is being dropped, thus its initial velocity will be 0;

• Initial velocity(u) = 0 m/s

Finally, the most imp point to understand is that, the height of tree will be just the distance covered by arecanut to reach the ground.

• Height of tree = distance covered by arecanut

From the top, whatever we have is t,a & u abd we need to find the distance covered(s), so by using 1rd equation of motion;

• v = u + at
• v = 0 + 10 × 1
• v = 10 m/s

As,we have velocity now, so by using 3rd equation of motion;

• v² - u² = 2as
• (10)² - (0)² = 2 × 10 × s
• 100 = 20s
• 100/20 = s
• 5 m = s

Therefore, at last the height of the tree is 5 metre.

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