A stone dropped from the top of a cliff reaches its foot in 4seconds claculate height of the cliff and the velocity with which the stone reaches the ground (g=9m/s).
Answers
Answer:
Velocity = 39.2 m/s
Height of the cliff = 78.4 m
Step-by-step explanation:
Given:
- Time taken = 4 s
- Initial velocity = 0 m/s
- Acceleration = 9.8 m/s²
To Find:
- Final velocity
- Height of the cliff
Concept:
Here we have to first find the acceleration we have to substitute the given data in the first equation of motion. For finding the distance travelled/height of the cliff, we have to use the second or third equation of motion.
Solution:
First we have to find the final velocity of the stone.
By the first equation of motion,
v = u + at
where v = final velocity
u = initial velocity
a = accleration
t = time taken
Here a = g = acceleration due to gravity
Substituting the data,
v = 0 + 9.8 × 4
v = 39.2 m/s
Hence velocity of the stone is 39.2 m/s.
Now we have to find the height of the cliff
Here height of the cliff is the distance travelled by the stone.
By the third equation of motion,
v² - u² = 2as
where v = final velocity
u = initial velocity
a = acceleration
s = distance travelled
Substitute the data,
39.2² - 0²= 2 × 9.8 × s
1536.64 = 19.6 s
s = 1536.64/19.6
s = 78.4
Hence the height of the cliff is 78.4 m.