Question

30.
A ball is projected vertically. upwards. At a height hi, its kinetic energy is 4 times as its
potential energy. At a height hz its potential energy is 4 times as its kinetic energy.
The relation between h, and he is
JA) 4h = hz (6) 4h2 = h. (c) 2h4 = hz (d) 2h₂ = hi​

Answer 1

Question:

A ball is projected vertically upwards. At a height $h_1,$ its kinetic energy is 4 times as its potential energy. And at a height $h_2,$ its potential energy is 4 times as its kinetic energy. Find the relation between $h_1$ and $h_2.$

Solution:

At the height $h_1,$ the velocity is given by the third kinematic equation as,

$(v_1)^2=u^2-2gh_1$

Then,

$K_1=4U_1\\\\\\\dfrac {1}{2}m(v_1)^2=4mgh_1\\\\\\\dfrac {1}{2}(u^2-2gh_1)=4gh_1\\\\\\u^2-2gh_1=8gh_1\\\\\\h_1=\dfrac {u^2}{10g}\quad\longrightarrow\quad(1)$

And, at the height $h_2,$ the velocity, similarly, is given by,

$(v_2)^2=u^2-2gh_2$

Initial velocity is the same for both cases.

Then,

$4K_1=U_1\\\\\\4\cdot\dfrac {1}{2}m(v_2)^2=mgh_2\\\\\\2(u^2-2gh_2)=gh_2\\\\\\2u^2-4gh_2=gh_2\\\\\\h_2=\dfrac {2u^2}{5g}\quad\longrightarrow\quad(2)$

Okay, now dividing (1) by (2), we get,

$\dfrac {h_1}{h_2}=\dfrac {1\times5}{10\times2}\\\\\\\dfrac {h_1}{h_2}=\dfrac {1}{4}\\\\\\\large\boxed {\mathbf{4h_1=h_2}}$