Question

A ball of mass 0.20 kg is thrown vertically upwards with an initial velocity of 20 metre per second -1 calculate the maximum potential energy it gains as it goes up​

Answer 1

Given

A ball of mass 0.2 kg is thrown vertically upwards with an initial velocity of 20 m/s

To Find

Maximum potential energy

Knowledge Required

$\bf \pink{\bigstar\ \; P=mgh}$

where ,

• P denotes potential energy
• m denotes mass
• g denotes gravity
• h denotes height

Height is not given here , so we need to apply equation of kinematics for finding height .

$\bf \green{\bigstar\ \; v^2-u^2=2as}$

where ,

• v denotes final velocity
• u denotes initial velocity
• a denotes gravity
• s denotes height

Solution

A body will gains maximum potential energy , when it is in maximum height . Lets find height ,

• m = 0.2 kg
• u = 20 m/s
• a = - 10 m/s²

Thrown against gravity

• v = 0 m/s

Goes to rest at extreme position

Apply 3rd equation of motion ,

$\bf \red{v^2-u^2=2as}\\\\\to \rm (0)^2-(20)^2=2(-10)h\\\\\to \rm -400=-20h\\\\\to \bf h=20\ m$

So , Maximum height = 20 m

Apply formula for potential energy ,

$\bf P=mgh\\\\\to \rm P=(0.2)(10)(20)\\\\\to \bf P=40\ J$

So , Maximum potential energy = 40 J

Answer 2

Given :-

• A ball of mass 0.20 kg is thrown vertically upwards with an initial velocity of 20 metre per second -1

To Calculate :-

• The maximum potential energy it gains as it goes up

Solution :-

We have,

• m = 0.2 kg
• u = 20 m/s
• a = - 10 m/s²
• v = 0 m/s

We know that,

$\boxed{\sf v^{2} - u^{2} = 2as} \: \: (\bf 3rd \: equation \: of \: motion)$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance covered

Substituting the values we get,

$:\implies\sf(0)^{2} - (20)^{2} = 2 \: (-10) \: h$

$:\implies\sf- 400 = - 20h$

$:\implies\sf20h = 400$

$:\implies\sf h = \dfrac{\cancel{400}}{\cancel{20}}$

$:\implies\sf h = 20 \: m$

$\green{\therefore\sf \: Maximum \: height = 20\:m.}$

★ We need to find the maximum potential energy it gains as it goes up :-

We know that,

$\boxed{\sf P = mgh} \: \: ( \bf Potential \: Energy)$

Substituting the values we get,

$:\implies\sf P = (0.2)(10)(20)$

$:\implies\sf P = 40\:J$

$\green{\therefore\sf Maximum \: potential \: energy = 40\:J.}$

______________________________

Additional Information :-

★ 1st equation of motion :-

$\boxed{\sf v = u + at}$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time taken

★ 2nd equation of motion :-

$\boxed{\sf s = ut + \dfrac{1}{2} at^{2}}$

Where,

• u = Initial Velocity
• t = Time taken
• a = Acceleration
• s = Distance covered

______________________________