Consider a projectile being launched at an initial velocity v0 in a direction making an angle θ with the horizontal. We assume that air resistance is negligible and the only force acting on the object is the force of gravity with acceleration g = 9.8 m/s2. Also an interactive html 5 applet may be used to better understand the projectile equations. Projectile problems with solutions are also included in this site.
Velocity, Acceleration and Displacement of the Projectile
The initial velocity V0 being a vector quantity, has two components:
V0x and V0y given by
V0x = V0 cos(θ)
V0y = V0 sin(θ)
The acceleration A is a also a vector with two components Ax and Ay given by
Ax = 0 and Ay = - g = - 9.8 m/s2
Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant and is given by
Vx = V0 cos(θ)
Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is given by
Vy = V0 sin(θ) - g t
Along the x axis the velocity Vxis constant and therefore the component x of the displacement is given by
x = V0 cos(θ) t
Along the y axis, the motion is that of a uniform acceleration type and the y component of the displacement is given by
y = V0 sin(θ) t - (1/2) g t2
Shape of the Trajectory of the Projectile
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 \( t \) by \( \dfrac{x}{V_0 cos(\theta)} \) in the above expression of y to obtain
\[ y = \dfrac{V_0 \sin(\theta) x}{V_0 \cos(\theta)} - (1/2) g \left( \dfrac{x}{V_0 \cos(\theta)} \right)^2 \]
Simplify
\[ y = - \dfrac{g \; x^2}{2( V_0 \cos(\theta))^2} + x \tan(\theta) \]
The above equation is the path of projectile which is a parabola of the form
\( y = A x^2 + B x \)
where \( A = - \dfrac{g}{2( V_0 \cos(\theta))^2} \) and \( B = \tan(\theta) \)
Time of Flight of the Projectile
The time of flight is the time taken for the projectile to go from point
A to point C (see figure above).
It is calculated by setting y = 0 (y = 0 at point C) and solve for t
y = V0 sin(θ) t - (1/2) g t2 = 0
Factor t out in the above equation
t(V0 sin(θ) - (1/2) g t) = 0
Two solutions:
t = 0 (correspond to point A)
and
t = 2 V0 sin(θ) / g (correspond to point C)
Hence the time of flight = 2 V0 sin(θ) / g
Maximum Height of the Projectile (corresponding to point B in figure above)
At point B in the figure above, the projectile is momentarily horizontal and therefore the vertical component of its velocity is equal to zero. Hence
Vy = V0 sin(θ) - g t = 0
Solve for t to obtain
t = V0 sin(θ) / g (Note: this is half the time of flight because of the symmetry of the parabola)
Substitute t by V0 sin(θ) / g in the expression of y, we obtain the maximum height
\( H = \dfrac{V_0 \sin(\theta) V_0 sin(\theta)}{g} - (1/2) \dfrac{}{} g \left( \dfrac{V_0 sin(\theta)}{g} \right)^2 = \dfrac{(V_0 sin(\theta))^2}{2 g} \)
Horizontal Range of a Projectile (distance AC in the figure above)
Distance AC which is the horizontal range is equal to x when t is equal to the time of flight 2 V0 sin(θ) / g obtained above. Hence
range AC = x = V0 cos(θ) t at t = time of flight = 2 V0 sin(θ) / g
Substitute t by 2 V0 sin(θ) / g and simplify to obtain the range AC
AC \( = V_0 cos(\theta) 2 V_0 sin(\theta) / g = V_0^2 sin(2\theta) / g \)
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