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:

ymax

=

- voy2 2 ay

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

ymax

=

vo2sin2(θ) 2 g

where g = 9.8 m/s²

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.