A particle is thrown from a building height h in upward direction with speed 10m/s.If particle during returning just missis the top of building and falls down.It takes time 10s to reach the ground. Find h.
Answers
Answer
- Height of the building = 400 m
Given
- Initial Velocity , u = 10 m/s
- Total time , t = 10 s
To Find
- Height of the ground , h
Concept Used
We need to apply equations of motion .
- v = u + at
- s = ut + ¹/₂ at²
- v² - u² = 2as
Solution
Find attachment for diagrammatic view of whole situation .
AB = BC
Height of the building , h = s" - s'
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Case - 1 : AB
Initial velocity , u = 10 m/s
Final velocity , v = 0 m/s [ Goes to rest at top position ]
Acceleration , a = g = - 10 m/s² [ Against the gravity ]
Time , t' = ? s
Distance , s' = ? m
Apply 3rd equation of motion .
⇒ v² - u² = 2as
⇒ 0² - 10² = 2 (-10) s'
⇒ - 100 = - 20 s'
⇒ s' = 5 m
Apply 1st equation of motion .
⇒ v = u + at
⇒ 0 = 10 + (-10)t'
⇒ 10t' = 10
⇒ t' = 1 s
__________________________
Given total time , t = 10 s
⇒ time for BD , t" = t - t'
⇒ t" = 10 - 1
⇒ t" = 9 s
Case - 2 : BD
Initial velocity , u = 0 m/s [ Starts from rest ]
Time , t" = 9 s
Acceleration , g = 10 m/s²
Distance , s" = ? m
Apply 2nd equation of motion .
⇒ s = ut + ¹/₂ at²
⇒ s" = (0)t" + ¹/₂ g(t")²
⇒ s" = ¹/₂ (10)(9)²
⇒ s" = 5 × 81
⇒ s" = 405 m
So , Height of the building , h = s" - s'
⇒ h = 405 - 5
⇒ h = 400 m
