Projectile Motion formula or equations for parabolic path, height, range, & time

In the next section, we will list down the Projectile Motion Formulas or equations. (Projectile trajectory equation & other formulas like maximum height and horizontal range of the projectile, time of flight, etc.)

In another page on this site, Find the derivation of Projectile equations – parabolic path, max height, range, & time of flight.

Projectile motion formula or equations

Projectile motion is the motion of a projected object in flight which is a result of two separate simultaneously occurring components of motion. One component of motion is without any acceleration along a horizontal direction and the other is along the vertical direction with constant acceleration due to the force of gravity (considering air resistance as negligible).

The initial velocity component along X-axis = V_0x = V₀ cosθ and the initial velocity component along Y-axis = V_0y = V₀sinθ.
(Air resistance is taken as negligible). g is the acceleration due to gravity.

Initial Horizontal Velocity	V0x = V0 cosθ [along X axis]
Horizontal velocity	VX = V0x = V0 cosθ [ uniform horizontal velocity]
Initial Vertical velocity	V0y = V0sinθ [along Y axis]
Vertical velocity	Vy = V0sinθ  – g t [ acceleration = -g]
Horizontal displacement	x= V0x.t = (V0 cosθ). t
Vertical displacement	y= (V0sinθ ).t  – (1/2) g t2
Motion Path equation (parabola equation):	y = (tanθ) x –   (1/2) g . x2/(V0 cosθ)2 [Parabolic]
Time to reach max height:	tmax= (V0sinθ )/g
Total time of flight for a projectile	Ttot = 2(V0sinθ )/g
Maximum height reached:	Hmax = ( V0sinθ )2/(2 g)
Horizontal range of a projectile:	R = (V02 sin2θ )/ g
Maximum possible horizontal range:	Rmax = V02 / g

Projectile Motion Solved Example

A ball is thrown at a speed of 28 m/s in a direction 30 degrees above the horizontal. Calculate (a) the maximum height (b) the time taken by the ball to return to the same level (c) the distance from the thrower to the point where the ball returns to the same level.

Solution:
V₀ = 28 m/s, θ = 30 degrees
(a) The maximum height = H_max = ( V₀sinθ )²/(2 g) = (28 Sin 30)^2 / (2×9.8) m = 10 m

(b) the time taken by the ball to return to the same level = Total time = T_tot = 2(V₀sinθ )/g = (2x28xsin 30) /9.8 = 2.9 s

(c) the distance from the thrower to the point where the ball returns to the same level = Horizontal range of a projectile = R = (V₀² sin2θ )/ g = (28x28xsin 60)/9.8 =69 m