Question

An areca nut took 1 second to reach the ground from the tree what will be the approximate hight of the arecanut tree? Which equation is used here

Answer 1

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

Question

An areca nut took 1 is the Ground from the tree water tap roots made height of the arecaunt tree ?

Answer

what is given here ?

• it is given your the time taken is one second
• the acceleration due to gravity is 9.8 metre per second square
• the final velocity is zero metre per second

What we have to find here ?

• we have to find here the formula which is to be used
• and the approximate height of the tree
• we also have to find here the initial velocity then only we can find the height of the tree

Formula to be used ?

• the first equation of motion

V=U+At

• the third equation of motion

V^2=U^2+2AS

Solution

first of all we have to find here the initial velocity so to find the initial velocity we have to use the first equation of motion

v=u+at

=> now substituting the values in their places .

=> 0=U +1×(-9.8)

=> 0=U-9.8

=> 0 = -9.8

=> -U=-9.8

=> U = 9.8

Hence , we have the initial velocity 9.8 .

now we have to find the height of the tree so we will use the third equation of motion .

V^2=U^2+2as

=> putting the values in their places

0^2=9.8^2+2(-9.8)×S

=> 0 = 96.04 +(-19.6)×s

=> -96.04 =-19.6× S

=> S=-96.04/-19.6

=> S= 4.9 m

Hence , the height of the tree is 4.9 m .