A stone is dropped from the tree. (Note: g=10m/s)
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?
Answers
Answer:
Correct option is B)
Distance (s) =20 m
Initial velocity =0 m/s
g=10 m/s
2
By using the formula:
v
2
=u
2
+2as
v
2
=0
2
+2×10×20
v
2
=400
v
2
=
400
v=20m/s
For calculating time:
V=u+at
20=0+10×t
t=
10
20
t=2 sec
The the stone will hit the ground with a velocity of 20 m/s in sec.
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