Consider a projectile that is launched from ground level at a particular angle. The launch point is the origin, and the motion stops when the projectile returns to ground level.

In other words:

• x_o = 0
• y_o = 0
• y = 0

The only acceleration is the acceleration due to gravity, and the positive directions are +x = right and +y = up.

Determine general equations that give, under the constraints above:

• the maximum height
• the time of flight
• the range (maximum horizontal distance)

The maximum height, y_max, can be found from the equation:

v_y ² = v_oy² + 2 a_y (y - y_o)

y_o = 0, and, when the projectile is at the maximum height, v_y = 0.

Solving the equation for y_max gives:

y_max = - v_oy² /(2 a_y)

Plugging in v_oy = v_o sin(θ) and a_y = -g, gives:

y_max = v_o²sin²(θ) /(2 g)

where g = 9.8 m/s²

Note that the maximum height is determined solely by the initial velocity in the y direction and the acceleration due to gravity. It's not affected by what's happening in the x direction.

To find the time of flight, determine the time the projectile takes to reach maximum height. The time of flight is just double the maximum-height time.

Start with the equation:

v_y = v_oy + a_y t

At maximum height, v_y = 0.

The time to reach maximum height is t_1/2 = - v_oy / a_y

Time of flight is t = 2t_1/2 = - 2v_oy / a_y

Plugging in v_oy = v_o sin(θ) and a_y = -g, gives:

Time of flight is t = 2 v_o sin(θ) / g

where g = 9.8 m/s²

The time of flight is also determined solely by the initial velocity in the y direction and the acceleration due to gravity.

The range is the horizontal distance traveled before the projectile hits the ground.

The distance covered in the x-direction is simply:

x = v_ox t

To find the range, use t as the time of flight, which we just calculated to be:

t = 2 v_o sin(θ) / g

Plugging in v_ox = v_o cos(θ) gives:

x = 2 v_o² sin(θ) cos(θ) / g

Using the trig. identity 2 sin(θ) cos(θ) = sin(2θ) this can be written as:

x = v_o² sin(2θ) / g