Question

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 = 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.

Answer 1

0.20m

0.46m

0.82m

1.28m

1.84m

e d g e n u i t y 2020

Step-by-step explanation:

Answer 2

Hello!

The answers are:

Height to achieve a speed of 2 m/s = 0.20m

Height to achieve a speed of 3 m/s = 0.46m

Height to achieve a speed of 4 m/s = 0.82m

Height to achieve a speed of 5 m/s = 1.28m

Height to achieve a speed of 6 m/s = 1.84m

Why?

Since an object in free fall is accelerated by gravity, the speed will increase according to the distance where the object is dropped. Gravitational acceleration is equal to 9.8 m/s2, it means that the speed will be increased by 9.8 m/s each second.

We are asked to find the heights necessary to drop the bottle to achieve the given speeds, so:

Speed of 2 m/s calculations:

`Ht_(2m/s) =(v^(2) )/(2g)=((2(m)/(s))^(2) )/(2*9.8(m)/(s^(2)))=(4(m^(2) )/(s^(2) ) )/(19.6(m)/(s^(2)))=0.20m`

Speed of 3 m/s calculations:

`Ht_(3m/s) =(v^(2) )/(2g)=((3(m)/(s))^(2) )/(2*9.8(m)/(s^(2)))=(9(m^(2) )/(s^(2) ) )/(19.6(m)/(s^(2)))=0.46m`

Speed of 4 m/s calculations:

`Ht_(4m/s) =(v^(2) )/(2g)=((4(m)/(s))^(2) )/(2*9.8(m)/(s^(2)))=(16(m^(2) )/(s^(2) ) )/(19.6(m)/(s^(2)))=0.82m`

Speed of 5 m/s calculations:

`Ht_(5m/s) =(v^(2) )/(2g)=((5(m)/(s))^(2) )/(2*9.8(m)/(s^(2)))=(25(m^(2) )/(s^(2) ) )/(19.6(m)/(s^(2)))=1.28m`

Speed of 6 m/s calculations:

`Ht_(6m/s) =(v^(2) )/(2g)=((6(m)/(s))^(2) )/(2*9.8(m)/(s^(2)))=(36(m^(2) )/(s^(2) ) )/(19.6(m)/(s^(2)))=1.84m`

Have a nice day!