The value of the variable that makes the expression on LHS and RHS equal is called the________of an equation. solution degree power equality About the author Ruby
Answer: Given:- Initial velocity ,u = 5m/s Final velocity ,v = 0m/s Acceleration due to gravity ,g= 10m/s² To Find:- Maximum height attained by stone ,h Total time taken to reach maximum height ,t Solution:- ⠀⠀⠀⠀⠀According to the Question It is given that the stone is thrown vertically upward direction . So the acceleration due to gravity on stone is negative. → g = -10m/s² Now, firstly we calculate the maximum height attained by the stone . Using 3rd equation of motion v² = u²+ 2gh where, v denote final velocity u denote initial velocity g denote acceleration due to gravity h denote maximum height attained by stone Substitute the value we get :\implies:⟹ 0² = 5² + 2(-10) × h :\implies:⟹ 0 = 25 -20h :\implies:⟹ -25 = -20h :\implies:⟹ 25 = 20h :\implies:⟹ 25/20 = h :\implies:⟹ h = 25/2 :\implies:⟹ h = 1.25 m Hence, the maximum height attained by stone is 1.25 metres. Now , calculating the time taken to reach maximum height . Using 1st equation of motion v = u + gt Substitute the value we get :\implies:⟹ 0 = 5 + (-10) × t :\implies:⟹ -5 = -10×t :\implies:⟹ 5 = 10 × t :\implies:⟹ 5/10 = t :\implies:⟹ t = 5/10 :\implies:⟹ t = 0.5s Hence, the time taken to reach maximum height is 0.5 second . _________________________________ Reply
Answer:
Given:-
Initial velocity ,u = 5m/s
Final velocity ,v = 0m/s
Acceleration due to gravity ,g= 10m/s²
To Find:-
Maximum height attained by stone ,h
Total time taken to reach maximum height ,t
Solution:-
⠀⠀⠀⠀⠀According to the Question
It is given that the stone is thrown vertically upward direction . So the acceleration due to gravity on stone is negative.
→ g = -10m/s²
Now, firstly we calculate the maximum height attained by the stone . Using 3rd equation of motion
v² = u²+ 2gh
where,
v denote final velocity
u denote initial velocity
g denote acceleration due to gravity
h denote maximum height attained by stone
Substitute the value we get
:\implies:⟹ 0² = 5² + 2(-10) × h
:\implies:⟹ 0 = 25 -20h
:\implies:⟹ -25 = -20h
:\implies:⟹ 25 = 20h
:\implies:⟹ 25/20 = h
:\implies:⟹ h = 25/2
:\implies:⟹ h = 1.25 m
Hence, the maximum height attained by stone is 1.25 metres.
Now , calculating the time taken to reach maximum height .
Using 1st equation of motion
v = u + gt
Substitute the value we get
:\implies:⟹ 0 = 5 + (-10) × t
:\implies:⟹ -5 = -10×t
:\implies:⟹ 5 = 10 × t
:\implies:⟹ 5/10 = t
:\implies:⟹ t = 5/10
:\implies:⟹ t = 0.5s
Hence, the time taken to reach maximum height is 0.5 second .
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