An online calculator to calculate the maximum height, range, time of flight, initial angle and the path of a projectile. The projectile equations and parameters used in this calculator are decribed below.

## 1 - Projectile Motion Calculator and Solver Given Initial Velocity, Angle and Height

Enter the initial velocity V_{0}in meters per second (m/s), the initial andgle θ in degrees and the initial height y

_{0}in meters (m) as positive real numbers and press "Calculate". The outputs are the maximum height, the time of flight, the range and the equation of the path of the form \( y = A x^2 + B x + C\).

## 2 - Projectile Motion Calculator and Solver Given Range, Initial Velocity, and Height

Enter the range in meters, the initial velocity V_{0}in meters per second and the initial height y

_{0}in meters as positive real numbers and press "Calculate". The outputs are the initial angle needed to produce the range desired, the maximum height, the time of flight, the range and the equation of the path of the form \( y = A x^2 + B x + C\) given V

_{0}and y

_{0}.

## Projectile Equations used in the Calculator and Solver

The vector initial velocity has two components:
V_{0x} and V_{0y} given by:

V_{0x} = V_{0} cos(θ)

V_{0y} = V_{0} sin(θ)

The vector acceleration A has two components A _{x} and A _{y} given by: (acceleration along the y axis only)

A_{x} = 0 and A_{y} = - g = - 9.8 m/s^{2}

At time t, the velocity has two components given by

V_{x} = V_{0} cos(θ)
and
V_{y} = V_{0} sin(θ) - g t

The displacement is a vector with the components x and y given by:

x = V_{0} cos(θ) t
and
y = y_{0} + V_{0} sin(θ) t - (1/2) g t^{2}

The time T_{m} at which y is maximum is at the vertex of y = y_{0} + V_{0} sin(θ) t - (1/2) g t^{2} and is given by

T_{m} = V_{0} sin(θ) / g

Hence the maximum height y_{max} reached by the projectile is given by

y_{max} = y(T_{m}) = y_{0} + V_{0} sin(θ) T_{m} - (1/2) g (T_{m})^{2}

The time of flight T_{f} is found by solving the equation

y_{0} + V_{0} sin(θ) t - (1/2) g t^{2} = 0

for t and taking the largest positive solution.

The shape of the trajectory followed by the projectile is found as follows

Solve the formula \( \; x = V_0 cos(\theta) t \; \) for \( t \) to obtain
\[ t = \dfrac{x}{V_0 cos(\theta)} \]

Substitute for t in y and simplify to obtain

\[ y = - \dfrac{g \; x^2}{2( V_0 \cos(\theta))^2} + x \tan(\theta) + y_0\]

The equation of the path of the projectile is a parabola of the form

\( y = A x^2 + B x + C\)

Horizontal Range = \( x(T_f) = V_0 cos(\theta) T_f \)

## More References and Links

Projectile Equations with ExplanationsInteractive Simulation of Projectile

Projectile problems with solutions .