Question

Find the time taken by vertically projected body to reach 3/4th of maximum height.
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Answer 1

Let initial velocity of the body be u.

We know the acceleration acting on a body projected vertically upward is $\sf{-g.}$

Let the maximum height be H at which the body has no final velocity.

$\longrightarrow\sf{v=0}$

By third equation of motion,

$\longrightarrow\sf{u^2-2gH=0}$

$\longrightarrow\sf{u^2=2gH}$

$\longrightarrow\sf{H=\dfrac{u^2}{2g}\quad\quad\dots(1)}$

By second equation of motion, the distance travelled by the body to reach $\sf{\left(\dfrac{3}{4}\right)^{th}}$ of its maximum height is,

$\longrightarrow\sf{\dfrac{3H}{4}=ut-\dfrac{1}{2}\,gt^2\quad\quad\dots(2)}$

where $\sf{t}$ is the time taken to travel this distance.

Substituting (1) in (2),

$\longrightarrow\sf{\dfrac{3u^2}{8g}=ut-\dfrac{1}{2}\,gt^2}$

$\longrightarrow\sf{\dfrac{1}{2}\,gt^2-ut+\dfrac{3u^2}{8g}=0}$

$\longrightarrow\sf{4g^2t^2-8ugt+3u^2=0}$

$\longrightarrow\sf{t=\dfrac{8ug\pm\sqrt{64u^2g^2-48u^2g^2}}{8g^2}}$

$\longrightarrow\sf{t=\dfrac{8ug\pm\sqrt{16u^2g^2}}{8g^2}}$

$\longrightarrow\sf{t=\dfrac{8ug\pm4ug}{8g^2}}$

$\longrightarrow\sf{t=\dfrac{8u+4u}{8g}\quad OR\quad t=\dfrac{8u-4u}{8g}}$

$\longrightarrow\sf{t=\dfrac{12u}{8g}\quad OR\quad t=\dfrac{4u}{8g}}$

$\longrightarrow\sf{t=\dfrac{3u}{2g}\quad OR\quad t=\dfrac{u}{2g}}$

The time $\sf{t=\dfrac{3u}{2g}}$ is attained when the body returns back to the point of projection. So we consider,

$\longrightarrow\sf{\underline{\underline{t=\dfrac{u}{2g}}}\quad\quad\dots(3)}$

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We know at maximum height, the body has no velocity.

$\longrightarrow\sf{v=0}$

Let $\sf{T}$ be the time taken to reach the maximum height. Then by first equation of motion.

$\longrightarrow\sf{u-gT=0}$

$\longrightarrow\sf{u=gT}$

$\longrightarrow\sf{T=\dfrac{u}{g}}$

$\longrightarrow\sf{T=2\times\dfrac{u}{2g}}$

From (3),

$\longrightarrow\sf{T=2t}$

$\longrightarrow\sf{t=\dfrac{T}{2}}$

∴ Time taken by the body to reach $\sf{\left(\dfrac{3}{4}\right)^{th}}$ of its maximum height is half the time taken to reach maximum height.

Answer 2

Answer:

John's age is 18

18 times 2

=36

36+3= 39

Explanation: