Question

Derive the formula for the range and
maximum height achieved by a projectile
thrown from the origin with initial
velocity u at an angel e to the horizontal,​

Answer 1

Solution:-

Here x and y - axis are in the direction show in the figure. axis c is along horizontal direction and axis y is vertically upward . Thus

$\rm \to \: \: u_x = ucos \alpha$

$\rm \to \: u_y = usin \alpha \: and \: a_x = 0$

$\rm \to \: a_y = - g$

At point A vertical component of velocity becomes zero I.e. Vy = 0 Substituting the proper value in

$\rm \to \: {v}^{2} _y = {u}^{2} _y + 2a_ys_y$

$\rm \to \: 0 = (usin \alpha ) {}^{2} + 2( - g)H$

$\rm \to \: H = \dfrac{ {u}^{2}sin {}^{2} \alpha }{2g}$

Hence derive

Definition

The point O is called the point of projection, The angle alpha is called angle of projection, the distance OB is called horizontal range

Answer 2

Answer:

$\huge\bf\blue{\underline{\underline{AnsWeR}}}$

Explanation:

Here x and y - axis are in the direction show in the figure. axis c is along horizontal direction and axis y is vertically upward . Thus

\rm \to \: \: u_x = ucos \alpha→u

x

=ucosα

\rm \to \: u_y = usin \alpha \: and \: a_x = 0→u

y

=usinαanda

x

=0

\rm \to \: a_y = - g→a

y

=−g

At point A vertical component of velocity becomes zero I.e. Vy = 0 Substituting the proper value in

\rm \to \: {v}^{2} _y = {u}^{2} _y + 2a_ys_y→v

y

2

=u

y

2

+2a

y

s

y

\rm \to \: 0 = (usin \alpha ) {}^{2} + 2( - g)H→0=(usinα)

2

+2(−g)H

\rm \to \: H = \dfrac{ {u}^{2}sin {}^{2} \alpha }{2g}→H=

2g

u

2

sin

2

α

Hence derive

Definition

The point O is called the point of projection, The angle alpha is called angle of projection, the distance OB is called horizontal range