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.
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
The shape of the trajectory followed by the projectile is found as follows
Solve the formula x=V0cos(θ)t for t to obtain
t=xV0cos(θ)
Substitute t by xV0cos(θ) in the above expression of y to obtain
y=V0sin(θ)xV0cos(θ)−(1/2)g(xV0cos(θ))2
Simplify
y=−gx22(V0cos(θ))2+xtan(θ)
The above equation is the path of projectile which is a parabola of the form
y=Ax2+Bx
where A=−g2(V0cos(θ))2 and B=tan(θ)
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
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=V0sin(θ)V0sin(θ)g−(1/2)g(V0sin(θ)g)2=(V0sin(θ))22g
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 =V0cos(θ)2V0sin(θ)/g=V20sin(2θ)/g