A body is dropped from a height of 320m. The acceleration due to the gravity is 10m/s2? (a)How long does it take to reach the ground? (b) What is the velocity with which it will strike the ground?
Answers
Explanation:
Given,
h = 320 m
g = 10 m/s²
u = 0
Using, v²−u² = 2gh
or v²−0 = 2×10×320 = 6400
or v = 80 m/s
This is the velocity with which the body will strike the ground.
Again, v = u+gt
or t = (v−u)/g = (80−0)/10 = 8 sec
Thus, (i) 8 sec, (ii) 80 m/s
Answer
a . Time , t = 8 seconds
b . Final velocity , v = 80 m/s
Given
- A body is dropped from a height of 320 m. The acceleration due to the gravity is 10 m/s²
To Find
a . Time
b . Final velocity
Concept Used
Equations of motion
- v = u + at
- s = ut + ¹/₂ at²
- v² - u² = 2as
Solution
Height , s = 320 m
Initial velocity , u = 0 m/s [ ∵ Dropped from height ]
Acceleration due to gravity , a = g = 10 m/s²
Time , t = ? s
Final velocity , v = ? m/s
Apply 2nd equation of motion .
⇒ s = ut + ¹/₂ at²
⇒ 320 = (0)t + ¹/₂ (10)t²
⇒ 320 = 5t²
⇒ t² = 64
⇒ t = 8 s
____________________
a . Time , t = 8 seconds
____________________
Apply 1st equation of motion .
⇒ v = u + at
⇒ v = 0 + (10)8
⇒ v = 80 m/s
_______________________
b . Final velocity , v = 80 m/s
_______________________