Question

1.
A particle is projected from horizontal plane (X-Z)
(where y-axis is along vertical) such that its velocity
at time t is v = ai + (b- ct)j. The horizontal range
of the particle is​

Answer 1

$\displaystyle\large\boxed {\sf {x=at}}$

To find horizontal range we only need to consider the horizontal motion.

So the initial velocity of the horizontal motion is,

• $\displaystyle\sf {v_x=a}$

(since the horizontal component of the velocity is not independent on time $\displaystyle\sf {t.}\!$)

Since no external force is acting horizontally, the acceleration,

• $\displaystyle\sf {\alpha_x=0}$

And the horizontal range,

• $\displaystyle\sf {s_x=x}$

Now, by the second kinematic equation, the horizontal range,

$\displaystyle\longrightarrow\sf {s_x=v_xt+\dfrac {1}{2}\alpha_xt^2}$

$\displaystyle\longrightarrow\sf {\underline {\underline {x=at}}}$