A stone is thrown vertically upwards with a speed of 20m/s . How high will it go before it begins to fall? (g= 9.8m/s²)
(with proper steps)
Answers
Answered by
223
Velocity at maximum height (v) = 0
Acceleration = -g
(-ve sign because it’s in the direction opposite to motion of stone)
v² - u² = 2aS
S = v² - u² / (2a)
= [0 - (20 m/s)²] / [2 × (-9.8 m/s²)]
= 20.40 m
It will reach a height of 20.40 m before it begins to fall.
Acceleration = -g
(-ve sign because it’s in the direction opposite to motion of stone)
v² - u² = 2aS
S = v² - u² / (2a)
= [0 - (20 m/s)²] / [2 × (-9.8 m/s²)]
= 20.40 m
It will reach a height of 20.40 m before it begins to fall.
rhearocx:
hey in the question they have asked to use g=9.8m/s². Then why did u use -9.8m/s² in ur answer?And Can we instead take u= -20m/s as it is in the opposite direction ?
Yeah before it begin to fall it won't be negative
I mentioned in my answer that why I took g = -9.8 m/s²
yeah..
OK
If you take u = -20 m/s then you should take g = +9.8m/s²
hmm that's what i wanted to know
thanks btw
You're welcome
Answered by
172
Hi !
Initial velocity = u = 20 m/s
Final velocity = v = 0 [ at maximum height , v = 0 ]
Acceleration due to gravity(g) in this case , is taken as negative.
This is because , when the direction of motion of object is opposite to "g" , then value of g is taken as -ve
hence ,
g = -9.8 m/s²
Let's use the formula :-
[h = height ]
v² -u² = 2gh
0² - 20² = 2*-9.8*h
-400 = -19.6h
h = -400/-19.6
= 20.408 m [ approximately ]
The stone will go 20.408 metres high before it begins to fall
Initial velocity = u = 20 m/s
Final velocity = v = 0 [ at maximum height , v = 0 ]
Acceleration due to gravity(g) in this case , is taken as negative.
This is because , when the direction of motion of object is opposite to "g" , then value of g is taken as -ve
hence ,
g = -9.8 m/s²
Let's use the formula :-
[h = height ]
v² -u² = 2gh
0² - 20² = 2*-9.8*h
-400 = -19.6h
h = -400/-19.6
= 20.408 m [ approximately ]
The stone will go 20.408 metres high before it begins to fall
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