- Enter the values of the two known variables in the text boxes

- Leave the text box empty for the variable you want to solve for

- Click on the calculate button.

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

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

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

- 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

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

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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