# Projectile Motion Calculator and Solver

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 V0 in meters per second (m/s), the initial andgle θ in degrees and the initial height y0 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 $$y = A x^2 + B x + C$$.

 Initial Velocity: V0 = 30m/s , Initial Angle: θ = 50° , Initial Height: y0 = 10m Decimal Places = 4 Maximum Height = meters Flight Time= seconds Range = meters Equation of the Path: y = x2 + x +

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

Enter the range in meters, the initial velocity V0in meters per second and the initial height y0 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 V0 and y0.

 Range = 50m, Initial Velocity: V0 = 30m/s , Initial Height: y0 = 10m Decimal Places = 4 Initial Angle = ° Maximum Height = meters Flight Time= seconds Equation of the Path:: y = x2 + x +

## Projectile Equations used in the Calculator and Solver

The vector initial velocity has two components: V0x and V0y given by:
V0x = V0 cos(θ)
V0y = V0 sin(θ)
The vector acceleration A has two components Ax and Ay given by: (acceleration along the y axis only)
Ax = 0 and Ay = - g = - 9.8 m/s2
At time t, the velocity has two components given by
Vx = V0 cos(θ) and Vy = V0 sin(θ) - g t
The displacement is a vector with the components x and y given by:
x = V0 cos(θ) t and y = y0 + V0 sin(θ) t - (1/2) g t2
The time Tm at which y is maximum is at the vertex of y = y0 + V0 sin(θ) t - (1/2) g t2 and is given by
Tm = V0 sin(θ) / g
Hence the maximum height ymax reached by the projectile is given by
ymax = y(Tm) = y0 + V0 sin(θ) Tm - (1/2) g (Tm)2
The time of flight Tf is found by solving the equation
y0 + V0 sin(θ) t - (1/2) g t2 = 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 = $$x(T_f) = V_0 cos(\theta) T_f$$