FIRST CORRECT ANSWER GETS BRAINLIEST You will drop the bottle/water mass so that it hits the lever at different speeds. Since an object in free fall is accelerated by gravity, you need to determine the heights necessary to drop the bottle to achieve the speeds of 2 m/s, 3 m/s, 4 m/s, 5 m/s, and 6 m/s. Use the equation Ht = v squared over 2 g. to calculate the height, where Ht is the height, v is the speed (velocity), and g is the gravitational acceleration of 9.8 m/s2.
Record these heights in Table B.
To achieve a speed of 2 m/s, the bottle must be dropped at
m.
To achieve a speed of 3 m/s, the bottle must be dropped at
m.
To achieve a speed of 4 m/s, the bottle must be dropped at
m.
To achieve a speed of 5 m/s, the bottle must be dropped at
m.
To achieve a speed of 6 m/s, the bottle must be dropped at
m.
Answers
Given :
Height necessary to drop the bottle to achieve the speeds of 2 m/s:
Hmax = u²/2g
where u= 2m/s
g= 9.8 m/s²
Hmax= 2 x2 / 2 x 9.8
=0.20 m
Height necessary to drop the bottle to achieve the speeds of 3 m/s:
Hmax = u²/2g
where u= 3m/s
g= 9.8 m/s²
Hmax= 3 x3 / 2 x 9.8
=0.46m
Height necessary to drop the bottle to achieve the speeds of 4 m/s:
Hmax = u²/2g
where u= 4m/s
g= 9.8 m/s²
Hmax= 4x4/ 2x 9.8
= 0.82m
Height necessary to drop the bottle to achieve the speeds of 5 m/s:
Hmax = u²/2g
where u= 5m/s
g= 9.8 m/s²
Hmax= 5x5/ 2 x9.8
=1.28m
Height necessary to drop the bottle to achieve the speeds of 6 m/s:
Hmax = u²/2g
where u= 6m/s
g= 9.8 m/s²
Hmax= 6x6/ 2x 9.8
= 18/9.8
=1.84m
Conclusion :
*********************************************************************************************
To acheive speed of Height must be
*********************************************************************************************
2m/s 0.20m
3m/s 0.46m
4m/s 0.82m
5m/s 1.28m
6m/s 1.84m
********************************************************************************************
Given :
Height necessary to drop the bottle to achieve the speeds of 2 m/s:
Hmax = u²/2g
where u= 2m/s
g= 9.8 m/s²
Hmax= 2 x2 / 2 x 9.8
=0.20 m
Height necessary to drop the bottle to achieve the speeds of 3 m/s:
Hmax = u²/2g
where u= 3m/s
g= 9.8 m/s²
Hmax= 3 x3 / 2 x 9.8
=0.46m
Height necessary to drop the bottle to achieve the speeds of 4 m/s:
Hmax = u²/2g
where u= 4m/s
g= 9.8 m/s²
Hmax= 4x4/ 2x 9.8
= 0.82m
Height necessary to drop the bottle to achieve the speeds of 5 m/s:
Hmax = u²/2g
where u= 5m/s
g= 9.8 m/s²
Hmax= 5x5/ 2 x9.8
=1.28m
Height necessary to drop the bottle to achieve the speeds of 6 m/s:
Hmax = u²/2g
where u= 6m/s
g= 9.8 m/s²
Hmax= 6x6/ 2x 9.8
= 18/9.8
=1.84m
Conclusion :
*********************************************************************************************
To acheive speed of Height must be
*********************************************************************************************
2m/s 0.20m
3m/s 0.46m
4m/s 0.82m
5m/s 1.28m
6m/s 1.84m
********************************************************************************************