Question

an object is projected with velocity v=(15i+20j). considering x along horizontal axis and y along vertical axis. find its velocity after 2 s.

Answer 1

an object is projected with velocity, $\bf{\vec{v}=15\hat{i}+20\hat{j}}$

we can separate velocity in x and y - components
so, $\bf{u_x=15}$
we know, acceleration of projectile in horizontal line is zero. e.g., $\bf{a_x=0}$
hence, velocity after 2 sec in horizontal direction.
$\bf{v_x=u_x+a_xt}\\\bf{v_x=u_x=15}$

similarly, $\bf{v_y=20}$
vertical direction , acceleration$\bf{a_y=-g}$ = -g[ negative sign shows acceleration acts vertically downward direction.]
now, velocity after 2 sect in vertical direction ,
$\bf{v_y=u_y+a_y.t}\\\bf{v_y=20-g\times2}\\\bf{v_y=20-10\times2=0}$

hence , velocity after 2 sec ,$\bf{\vec{V}=15\hat{i}+0\hat{j}}$

Answer 2

Here is your answer

Answer:

an object is projected with velocity, \bf{\vec{v}=15\hat{i}+20\hat{j}}

v

=15

i

^

+20

j

^

we can separate velocity in x and y - components

so, \bf{u_x=15}u

x

=15

we know, acceleration of projectile in horizontal line is zero. e.g., \bf{a_x=0}a

x

=0

hence, velocity after 2 sec in horizontal direction.

\begin{gathered}\bf{v_x=u_x+a_xt}\\\bf{v_x=u_x=15}\end{gathered}

v

x

=u

x

+a

x

t

v

x

=u

x

=15

similarly, \bf{v_y=20}v

y

=20

vertical direction , acceleration\bf{a_y=-g}a

y

=−g = -g[ negative sign shows acceleration acts vertically downward direction.]

now, velocity after 2 sect in vertical direction ,

\begin{gathered}\bf{v_y=u_y+a_y.t}\\\bf{v_y=20-g\times2}\\\bf{v_y=20-10\times2=0}\end{gathered}

v

y

=u

y

+a

y

.t

v

y

=20−g×2

v

y

=20−10×2=0

hence , velocity after 2 sec ,\bf{\vec{V}=15\hat{i}+0\hat{j}}

V

=15

i

^

+0

j

^