Question

HELP ASAP TIMED TEST

A snowball is dropped from a height of H and reaches the ground with a speed of V. If you want to double the speed that the snowball has when it reaches the ground, at what height should you drop it from?

a
2H
b
4H
c
√2H

d
8H
e
16H

Answer 1

Correct choice: b 4H

Step-by-step explanation:

Conservation of the mechanical energy

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

`\displaystyle K=(1)/(2)mv^2`

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

`U_1 = mgH`

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

`\displaystyle U_2=(1)/(2)mv^2`

Since the energy is conserved, U1=U2

`\displaystyle mgH=(1)/(2)mv^2 \qquad\qquad [1]`

For the speed to be double, we need to drop the snowball from a height H', and:

`\displaystyle mgH'=(1)/(2)m(2v)^2`

Operating:

`\displaystyle mgH'=4(1)/(2)m(v)^2 \qquad\qquad [2]`

Dividing [2] by [1]

`\displaystyle (mgH')/(mgH)=(4(1)/(2)m(v)^2)/((1)/(2)m(v)^2)`

Simplifying:

`\displaystyle (H')/(H)=4`

Thus:

H' = 4H

Correct choice: b 4H