Question

28. A stone is dropped from the tree. (Note: g=10m/s^2)

a) How long does it take to fall 20m?

b) How fast does it move at the end of that fall?

c) What is its acceleration after 1s and after 2s?

Answer 1

Stone takes 2 secs to fall 20 m and moves at 20 m/s at  the end of that fall.  Acceleration after 1s and after 2s is same i.e  g = 10 m/s²

Given:

• A stone is dropped from the tree
• g = 10 m/s²

To Find:

• a) How long does it take to fall 20m?
• b) How fast does it move at the end of that fall?
• c) What is its acceleration after 1s and after 2s?

Solution:

• S = ut + (1/2)at²
• v = u + at
• v² = u² + 2aS

Step 1:

Use S = ut + (1/2)at²  and substitute S = 20 , u = 0  a = g = 10   and calculate t

=> 20 = 0 + (1/2)(10)t²

=> 4 = t²

=> 2 = t    

Considering positive value only as time can not be negative)

Hence, It  takes 2 secs to fall 20 m

Step 2:

Use v = u + at  and substitute u = 0 , a = g = 10  and t = 2 and calculate v

v = 0 + 10 (2)

v = 20

Hence , it moves at 20 m/s at  the end of that fall

Step 3:

Acceleration after 1s and after 2s is same i.e  g = 10 m/s²

Stone takes 2 secs to fall 20 m and moves at 20 m/s at  the end of that fall.  Acceleration after 1s and after 2s is same i.e  g = 10 m/s²