A ball is thrown straight up into the air with a speed of 13 m/s. If the ball has a mass of 0.25 kg, how high does the ball go? Acceleration due to gravity is g=9.8m/s^2

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gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

The total energy, i.e. sum of kinetic and potential energy, is constant.

i.e. E = KE + PE

Initially, PE = 0 and KE = 1/2 mv^2

At maximum height, velocity=0, thus, KE = 0 and PE = mgh

Since, total energy is constant (KE converts to PE when the ball is rising),

therefore, KE = PE 

or, 1/2 mv^2 = mgh

or, h = v^2 /2g = 13^2 / (2x9.8) = 8.622 m

Hope this helps.

piyumie's profile pic

Piyumi | Student, Grade 12 | (Level 1) eNoter

Posted on

u =`13 ms^-1`

` `

v= `0`

g/a = - `9.8ms^-2`

s = `?`

`v^2 = u^2 + 2*-a*s`

`0 =` `13^2 + 2 * -9.8 * S`

`S = -169/19.6`

` `

`s = 8.6244`

`s = 8.62m`

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mangoes123's profile pic

mangoes123 | Student | (Level 1) eNoter

Posted on

Things to keep in mind:

`KE = 1/2mv^2`

`PE=mgh`

(KE is Kinetic Energy and PE is Potential Energy) 

So, to solve this, it's best to use energies. When the ball is first thrown is when its KE is at MAX so we put that on the left side, and when its reaches maximum height, its potential energy is GREATEST so we put the on the right side of the equation so:

`1/2mv^2=mgh` 

m is the mass  
g is the acceleration due to gravity 
v is the velocity
h is the final height 

 which is the same as 

Now, plug the values into the equation:

`1/2(.25)(13)^2=(.25)(9.8)h`

Then, solve.

`1/2(.25)(169)=(2.45)h`

`1/2(42.25)=(2.45)h`

`(21.125)=(2.45)h `

`21.125/2.45=h`

` 8.6224=h `

The unit is, obviously, meters. I chose to leave out the units while solving as it gets confusing.

I hope this helps!

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kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

Irrespective of mass, for any body when thrown upwards it goes to a height of S meters (when initial velocity "U " is given to the body).

The S value can be obtained from the standard formula i.e

`V^2 - U^2 = 2*a*S`

where, V is the final velocity

U is the initial velocity

a is the acceleration

S is the distance covered.

now while coming to the question given above, the ball is thrown upwards

so a = -g 

V = 0

u =13 m/s

so,

`V^2 - U^2 = 2*a*S`

=> `0^2 - U^2 = 2*(-g)*S`

=> `(- U^2) = 2*(-g)*S`

=> `S=(u^2)/(2*g)`

=`(13^2)/(2*(9.8))`

= 8.6224 m

so the ball went 8.6224 m high.

:)

Sources:

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