Question

a body is projected vertically up with velocity 19.6 then find 1.displacement in 2 seconds 2.velocity after 3 seconds 3.ratio of displacement in 1st and 2nd second respectively​

Answer 1

The initial velocity of the body is,

• $\sf{u=19.6\ m\,s^{-1}}$

Since the body is projected vertically upwards, the acceleration acting on the body is,

• $\sf{a=-g=-9.8\ m\,s^{-2}}$

By second equation of motion, the displacement in first 2 seconds is,

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

$\longrightarrow\sf{s=19.6\times2-\dfrac{1}{2}\times9.8\times2^2}$

$\longrightarrow\sf{s=39.2-19.6}$

$\longrightarrow\sf{\underline{\underline{s=19.6\ m}}}$

By first equation of motion, the velocity of the body after first 3 seconds is,

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

$\longrightarrow\sf{v=19.6-9.8\times3}$

$\longrightarrow\sf{v=19.6-29.4}$

$\longrightarrow\sf{\underline{\underline{v=-9.8\ m\,s^{-1}}}}$

The negative sign shows the velocity is acting downwards at this moment.

The displacement of the body in 1st second is,

$\longrightarrow\sf{s_1=ut+\dfrac{1}{2}\,at^2}$

$\longrightarrow\sf{s_1=19.6\times1-\dfrac{1}{2}\times9.8\times1^2}$

$\longrightarrow\sf{s_1=19.6-4.9}$

$\longrightarrow\sf{s_1=14.7\ m}$

The displacement in 2nd second is found earlier.

$\longrightarrow\sf{s_2=19.6\ m}$

Hence the ratio of the displacements in 1st and 2nd seconds is,

$\longrightarrow\sf{\dfrac{s_1}{s_2}=\dfrac{14.7}{19.6}}$

$\longrightarrow\sf{\underline{\underline{\dfrac{s_1}{s_2}=\dfrac{3}{4}}}}$