Vertical projectile motion

When we launch a projectile vertically upwards:

1. Its initial vertical velocity is positive (upwards).
2. It will decelerate at a constant rate due to the acceleration of gravity, which is negative (downwards) so decelerates the projectile.
3. At the maximum height, the vertical velocity will be zero because the acceleration due to gravity has fully decelerated the projectile to a velocity of 0ms^{-1}.
4. After reaching the maximum height, the projectile will start to fall back down, accelerating at the same rate due to gravity, but now with a negative velocity (downwards).
5. When the projectile hits the ground, it will have the same speed as when it was launched, but in the opposite direction (downwards). This is because the acceleration due to gravity is constant.

Equations

We can use the same equations for uniform acceleration to solve problems involving vertical projectile motion:

• v = u + at
• s=\frac{u+v}2\times t
• v^2 = u^2 + 2as
• s=ut+\frac12at^2

Known values

there are some values which we will always know when solving vertical projectile motion problems.

Values for the upwards part of the motion

• The acceleration (a) will always be -9.81ms^{-2} (the acceleration due to gravity, which is negative because it’s downwards).
• The initial velocity (u) will always be positive (since it’s launched upwards). It’s the velocity it was launched at.
• The final velocity (v) will always be 0ms^{-1} at the maximum height, because the projectile will have been fully decelerated by gravity, and, for an instant, will be completely stationary (no velocity).
• The displacement (s) will be the maximum height reached by the projectile (compared to the launch point). It’s positive as the projectile is going up.
• The time (t) will be the time taken to reach the maximum height.

Values for the downwards part of the motion

• The acceleration (a) will still be -9.81ms^{-2} (the acceleration due to gravity, which is negative because it’s downwards).
• The initial velocity (u) will always be 0ms^{-1} at the maximum height, because the projectile will have been fully decelerated by gravity, and, for an instant, will be completely stationary (no velocity).
• The final velocity (v) will always be negative (since it’s falling downwards). It’s the velocity it hits the ground with, and will have the same magnitude as the initial velocity (but negative) due to the constant acceleration of gravity.
• The displacement (s) will be the maximum height reached by the projectile (compared to the launch point), but negative, since the projectile is falling down.
• The time (t) will be the time taken to fall back down to the ground.