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

$R =\frac {u^2 sin 2 \theta}{g}$

Where $u$ is the velocity of launch

$\theta$ is the angle of launch

H and R is the Maximum Height and Range

Value of g is assumed at 10m/s

Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown with some initial velocity near the earth's surface, and it moves along a curved path under the action of gravity alone.

Maximum Height And Range is given by the below formula $H= \frac {u^2 sin^2 \theta}{2g}$

$R =\frac {u^2 sin 2 \theta}{g}$

Where

H -> Maximum Height

R- > Range

u -> initial Velocity

$\theta$ -> Angle projected

g -> Accleration due to gravity

A object is projected with velocity 2 m/s at an angle 45°.Find the Range and Maximum Height attained by the object in Projectile Motion

Given u=2 m/s, $\theta = 45^0$, g=10 m/s, H =? , R=?

Maximum Height And Range is given by the below formula

$H= \frac {u^2 sin^2 \theta}{2g} = \frac { 2^2 sin^2 45}{2 \times 10} = \frac {2}{20} = .1 \ m $

$R =\frac {u^2 sin 2 \theta}{g}= \frac { 2^2 sin 90}{10} = .4 \ m $

if u =40 m/s and H=25 m , Find the angle and range of the Projectile?

Given u=40 m/s, $\theta = ?$, g=10 m/s, H =25 m , R=?

Angle And Range is calculated as below ( $g=10 \ m/s^2$)

$\theta=sin^{-1} \sqrt {\frac {2gH}{u^2}}=sin^{-1} \sqrt {\frac {2 \times 10 \times 25}{40^2}}=33.98^0 $

$R =\frac {u^2 sin 2 \theta}{g}= 148.32 \ m $

Maximum Height And Range is calculated as below ( $g=10 \ m/s^2$)

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

$R =\frac {u^2 sin 2 \theta}{g}$

2. If initial velocity and Maximum height are given

Angle And Range is calculated as below ( $g=10 \ m/s^2$) $\theta=sin^{-1} \sqrt {\frac {2gH}{u^2}} $

$R =\frac {u^2 sin 2 \theta}{g}$

3. If initial velocity and Range are given

Angle And Maximum Height is calculated as below ( $g=10 \ m/s^2$)

$\theta=\frac {sin^{-1} \frac {gR}{u^2}}{2} $

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

4. if Maximum Height and Range are given

Angle And initial velocity is calculated as below ( $g=10 \ m/s^2$)

$\theta= tan^{-1} \frac {4H}{R}$

$u= \sqrt { \frac {gR}{sin 2 \theta}}$

5. if Angle and Range are given

initial velocity And Maximum Height is calculated as below ( $g=10 \ m/s^2$)

$u= \sqrt { \frac {gR}{sin 2 \theta}}$

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

6. if Angle and Maximum Height are given

initial velocity And Range is calculated as below ( $g=10 \ m/s^2$)

$u= \sqrt { \frac {2gH}{sin^2 \theta}}$

$R =\frac {u^2 sin 2 \theta}{g}$

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