Question

A ball having the potential energy of 1000 J
is dropped from a height of 100 m. At
approximate what height (in metres) will the
kinetic energy be 5 times the potential
energy at that height?​

Answer 1

Given:

A ball having the potential energy of 1000 J is dropped from a height of 100 m.

To find:

Height at which kinetic energy will be 5 times that of the potential energy.

Calculation:

Let the height from ground be x ;

$\rm{ \therefore \: kinetic \: energy = 5 \times (potential \: energy)}$

$\rm{ = > \: \dfrac{1}{2} m {v}^{2} = 5 \times (m \times g \times x)}$

$\rm{ = > \: \dfrac{1}{2} {v}^{2} = 5 \times ( g \times x)}$

$\rm{ = > \: \dfrac{1}{2} \bigg \{ {u}^{2} + 2g(h - x) \bigg \} = 5gx}$

$\rm{ = > \: \bigg \{ {u}^{2} + 2g(h - x) \bigg \} = 10gx}$

$\rm{ = > \: {u}^{2} + 2g(h - x)= 10gx}$

$\rm{ = > \: {u}^{2} + 2gh - 2gx= 10gx}$

$\rm{ = > \: {u}^{2} + 2gh= 12gx}$

$\rm{ = > \: {(0)}^{2} + 2gh= 12gx}$

$\rm{ = > \: 0 + 2gh= 12gx}$

$\rm{ = > \: 2gh= 12gx}$

$\rm{ = > \: 2h= 12x}$

$\rm{ = > \: h= 6x}$

$\rm{ = > \: x = \dfrac{h}{6} }$

$\rm{ = > \: x = \dfrac{100}{6} }$

$\rm{ = > \: x = 16.67 \: m }$

So, final answer is:

$\boxed{ \bf{ \: height = x = 16.67 \: m }}$

Answer 2

Given:

• A ball having the potential energy of 1000 J is dropped from a height of 100 m.

To find:

• Height at which kinetic energy will be 5 times that of the potential energy.

Calculation:

Let the height from ground be x .

Then,

$\rm{ \: kinetic \: energy = 5 \times (potential \: energy)}$

$\rm{ = > \: \dfrac{1}{2} m {v}^{2} = 5 \times (m \times g \times x)} </p><p>$

$\rm{ = > \: \dfrac{1}{2} {v}^{2} = 5 \times ( g \times x)}$

$\rm{ = > \: \dfrac{1}{2} \bigg \{ {u}^{2} + 2g(h - x) \bigg \} = 5gx}$

$\rm{ = > \: \bigg \{ {u}^{2} + 2g(h - x) \bigg \} = 10gx}$

$\rm{ = > \: {u}^{2} + 2g(h - x)= 10gx} \\ \\ \\ </p><p></p><p>\rm{ = > \: {u}^{2} + 2gh - 2gx= 10gx}\\ \\ \\ </p><p></p><p>\rm{ = > \: {u}^{2} + 2gh= 12gx}\\ \\ \\ </p><p></p><p>\rm{ = > \: {(0)}^{2} + 2gh= 12gx}\\ \\</p><p></p><p>\rm{ = > \: 0 + 2gh= 12gx}\\ \\</p><p></p><p>\rm{ = > \: 2gh= 12gx}\\ \\</p><p></p><p>\rm{ = > \: 2h= 12x}\\ \\</p><p></p><p>\rm{ = > \: h= 6x}\\ \\</p><p></p><p>\rm{ = > \: x = \dfrac{h}{6} }\\ \\</p><p> </p><p> </p><p></p><p>\rm{ = > \: x = \dfrac{100}{6} }\\ \\ </p><p> </p><p> </p><p></p><p>\rm{ = > \: x = 16.67 \: m }\\ \\ </p><p>=>x=16.67m$

Hence, required answer is:

$\implies \boxed{ \bf{\blue{ \: height = 16.67 \: m} }}$