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.
The formula used for solving the question is
$H= \frac {u^2 sin^2 \theta}{2g}$
$R =\frac {u^2 sin 2 \theta}{g}$
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/s2
Projectile Motion Calculator
What is Projectile Motion
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 is the Maximum height reached by projectile during the motion Range is horizontal distance covered by the projectile during the motion.
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
Example of Few questions where you can use this Projectile Motion calculator Question 1
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 Solution
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 $
Question 2
if u =40 m/s and H=25 m , Find the angle and range of the Projectile? Solution
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 $
How the Projectile Motion calculator works
1. If initial velocity and angle are given
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 ( $g=10 \ m/s^2$)
$u= \sqrt { \frac {2gH}{sin^2 \theta}}$
$R =\frac {u^2 sin 2 \theta}{g}$
