An object of mass m is dropped freely from a height of ‘h’ and it reaches to the ground in time ‘t’. then calculate(i) What is its speed on striking the ground?(ii) What is its average speed during time t?(iii) How high the object is being dropped above the ground?
Answers
Answer:
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Explanation:
Speed on striking the ground is √(2gh) , Average speed = h/t and The object is being dropped above the ground at height h
Given:
- Object of mass m dropped from height h
- Reached ground in Time t
To Find:
- (i) Speed on striking the ground
- (ii) Average speed during time t
- (iii) How high the object is being dropped above the ground ( Its already given height h)
Solution:
- TE = KE + PE
- TE = Total Energy
- KE = Kinetic Energy = (1/2)mv²
- PE = Potential Energy = mgh
- Average Speed = Total Distance / Time
Step 1:
Initially object is at height h and at rest (as dropped) so
TE = mgh + (1/2)m(0)²
TE = mgh
Step 2:
Finally object is at height 0 (ground) and speed v
TE = mg(0) + (1/2)m(v)²
TE = (1/2)m(v)²
Step 3:
Equate Total Energy
(1/2)m(v)² = mgh
=> (v)² = 2gh
=> v = √(2gh)
Hence speed on striking the ground is √(2gh)
Step 4:
Use Average Speed = Total Distance / Time and
substitute Total distance = h and Total time = t
Hence Average speed = h/t
Step 5:
The object is being dropped above the ground at height h is already given in the Question.
Speed on striking the ground is √(2gh) , Average speed = h/t and The object is being dropped above the ground at height h