# 1.4 Projectile motion types  (Page 4/5)

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## Projectile thrown at an angle with horizontal direction

There are two possibilities. The projectile can be projected up or down as shown in the figure here :

## Projectile thrown up at an angle with horizontal direction

The time of flight is determined by analyzing motion in vertical direction. The net displacement during the motion is equal to the elevation of point of projection above ground i.e. ${H}_{2}$ . To analyze the motion, we consider point of projection as origin, horizontal direction as x-axis and upward vertical direction as y-axis.

$y=-{H}_{2}={u}_{y}T-\frac{1}{2}g{T}^{2}$

Rearranging,

${T}^{2}-\left(\frac{2{u}_{y}}{g}\right)T+\left(\frac{2{H}_{2}}{g}\right)=0$

This is a quadratic equation in “T”. Solving we get two values of T, one of which gives the time of flight.

Horizontal range is given by analyzing motion in horizontal direction as :

$x=R={u}_{x}T$

While calculating maximum height, we can consider motion in two parts. The first part is the motion above the projection level. On the other hand second part is the projectile motion below projection level. The total height is equal to the sum of the magnitudes of vertical displacements :

$H=|{H}_{1}|+|{H}_{2}|$

Problem : A projectile is thrown from the top of a building 160 m high, at an angle of 30° with the horizontal at a speed of 40 m/s. Find (i) time of flight (ii) Horizontal distance covered at the end of journey and (iii) the maximum height of the projectile above the ground.

Solution : Unlike horizontal projection, the projectile has a vertical component of initial velocity. This vertical component is acting upwards, which causes the projectile to rise above the point of projection.

Here, we choose the point of projection as the origin and downward direction as the positive y – direction.

(i) Time of flight, T

Here,

$\begin{array}{l}{u}_{y}=u\mathrm{sin}\theta =-40\mathrm{sin}{30}^{0}=-20\phantom{\rule{2pt}{0ex}}m/s;\\ y=160\phantom{\rule{2pt}{0ex}}m;\\ g=10\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

Using equation, $y={u}_{y}t+\frac{1}{2}g{t}^{2}$ , we have :

$\begin{array}{l}⇒160=-20t+\frac{1}{2}10{t}^{2}\\ ⇒5{t}^{2}-20t-160=0\\ ⇒{t}^{2}-4t-32=0\\ ⇒{t}^{2}-8t+4t-32=0\\ ⇒t\left(t-8\right)+4\left(t-8\right)=0\\ ⇒t=-4\phantom{\rule{2pt}{0ex}}s\phantom{\rule{4pt}{0ex}}\mathrm{or}\phantom{\rule{4pt}{0ex}}t=8\phantom{\rule{2pt}{0ex}}s\end{array}$

Neglecting negative value of time, T = 8 s.

(ii) Horizontal distance, R

There is no acceleration in horizontal direction. Using equation for uniform motion,

$\begin{array}{l}x={u}_{x}T\end{array}$

Here,

$\begin{array}{l}{u}_{x}=u\mathrm{cos}\theta =40\mathrm{cos}{30}^{0}=20\sqrt{3}\phantom{\rule{2pt}{0ex}}m/s;\\ T=8\phantom{\rule{2pt}{0ex}}s\end{array}$

$\begin{array}{l}⇒x={u}_{x}T=20\sqrt{3}x8=160\sqrt{3}\phantom{\rule{2pt}{0ex}}m\end{array}$

(iii) Maximum height, H

The maximum height is the sum of the height of the building ( ${H}_{2}$ ) and the height attained by the projectile above the building ( ${H}_{1}$ ).

$\begin{array}{l}H={H}_{1}+{H}_{2}\end{array}$

We consider vertical motion to find the height attained by the projectile above the building ( ${H}_{1}$ ).

${H}_{1}=\frac{{u}^{2}{\mathrm{sin}}^{2}\theta }{2g}$

$⇒{H}_{1}=\frac{{\left(40\right)}^{2}{\mathrm{sin}}^{2}{30}^{0}}{2X10}$

$⇒{H}_{1}=\frac{{\left(40\right)}^{2}X\left(\frac{1}{4}\right)}{2X10}=20\phantom{\rule{1em}{0ex}}m$

Thus maximum height, H, is :

$\begin{array}{l}H={H}_{1}+{H}_{2}=20+160=180\phantom{\rule{2pt}{0ex}}m\end{array}$

## Projectile thrown down at an angle with horizontal direction

Projectile motion here is similar to that of projectile thrown horizontally. The only difference is that the projectile has a finite component of velocity in downward direction against zero vertical velocity.

For convenience, the point of projection is considered as origin of reference and the positive x and y directions of the coordinate system are considered in horizontal and vertically downward directions.

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