Question

A particle is projected from a point $\left( {0,1} \right)$ on $Y$ -axis (assume $+ Y$ direction vertically upwards) aiming towards a point$\left( {4,9} \right)$ . It fell on the ground on $X$ -axis in $1$ sec. Taking $g = 10m/{s^2}$ and all coordinates in meter.​ Find the $X$ coordinate where it fell.

Answer 1

The slope that is equal to the tangent of the angle of the projection has to be calculated from the given points. The speed at which the particle is projected is to be divided into two components along with horizontal and vertical directions at the projection angle. Using the equation of motion, the speed along the vertical direction will be found. From this value, the speed along the horizontal direction can be calculated. The x-coordinate can be calculated from this speed.

Formula used:
Slope = $\tan \theta = \dfrac{{{y_2} - {x_2}}}{{{y_1} - {x_1}}}$
Displacement for points on the y-axis, ${S_1} = {u_y}t - \dfrac{1}{2}g{t^2}$
Displacement for points on the x-axis , ${S_2} = {u_x}t$
${u_x}$ and ${u_y}$ are the components of the projection velocity $u$ along with horizontal and vertical directions.

Complete step by step solution:
The points are given, starts from $\left( {0,1} \right)$ and aiming to $\left( {4,9} \right)$
Slope = $\tan \theta = \dfrac{{9 - 1}}{{4 - 0}}$
$\tan \theta = \dfrac{8}{4} = 2$
Let, ${u_x}$ and ${u_y}$ are the components of the projection velocity $u$ along with horizontal and vertical directions.
${u_x} = u\cos \theta$
${u_y} = u\sin \theta$
$\theta$ is the projection angle.
The equation of motion of the particle vertically,
${S_1} = {u_y}t - \dfrac{1}{2}g{t^2}$
${S_1} = 0 - 1 = - 1$
$t = 1\sec$
$g = 10m/{s^2}$
$\therefore ( - 1) = \left( {u\sin \theta .1} \right) - \left( {\dfrac{1}{2}.10.1} \right)$
$\Rightarrow u\sin \theta = 5 - 1$
$\Rightarrow u\sin \theta = 4$
$\tan \theta$ can be written as, $\tan \theta = \dfrac{{u\sin \theta }}{{u\cos \theta }}$
$\Rightarrow 2 = \dfrac{4}{{u\cos \theta }}$
$\Rightarrow u\cos \theta = 2$
The equation of motion of the particle horizontally,
${S_2} = {u_x}t$
$\Rightarrow {S_2} = u\cos \theta t$
$\Rightarrow {S_2} = 2 \times 1$
$\Rightarrow {S_2} = 2$
Hence, the $X$ coordinate is $\left( {2,0} \right)$ where it fell.

Note: The component of the projection speed along the vertical direction is non-uniform. The motion of the particle is along with the acceleration due to gravity. But the component along with the horizontal direction is uniform since there is no gravitational acceleration.
When we want to find the maximum height of the particle, we have to deal with the vertical motion. And, during the calculation of the range of the particle, we have to deal with the horizontal motion.