Question

An aracanut took 1 s to reach the ground from the tree. What will be the approximate height of the tree? (g= 10 m/s2)​

Answer 1

Answer:

Given:-

• Time taken ,t = 1s

• Acceleration due to gravity ,g = 9.8 m/s²

• Final velocity ,v = 0m/s

To Find:-

• Height ,h

Solution:-

Here , we apply 1st & 3rd equation of motion to find the height of tree.

By using 1st equation of motion

• v = u+at

Substitute the value we get

→ 0 = u + 1×(-9.8)

→ 0 = u -9.8

→ -u = -9.8

→ u = 9.8

∴ The initial velocity of the nut is 9.8 m/s

Now , Using 3rd equation of motion

• v² = u² +2ah

Substitute the value we get

→ 0² = 9.8² + 2× (-9.8) × h

→ 0 = 96.04 + (-19.6) ×h

→ -96.04 = - 19.6 ×h

→ h = -96.04/19.6

→ h = 4.9 m

∴ The height of the areca but tree is 4.9 m .

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Answer 2

Answer:

The answer is 4.9 m

If my answer is correct please mark me as brainlist