Vertical projectile motion

When we launch a projectile vertically upwards:

Its initial vertical velocity is positive (upwards).
It will decelerate at a constant rate due to the acceleration of gravity, which is negative (downwards) so decelerates the projectile.
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}.
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).
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.

flashcards

Question	Answer
What happens to the initial vertical velocity of a projectile launched vertically upwards?	It is positive (upwards).
What happens to the vertical velocity of a projectile launched vertically upwards as it rises?	It decelerates at a constant rate due to the acceleration of gravity, which is negative (downwards).
What is the vertical velocity of a projectile at its maximum height?	It is zero, because the acceleration due to gravity has fully decelerated the projectile to a velocity of 0ms^{-1}.
What happens to a projectile after it reaches its maximum height?	It starts to fall back down, accelerating at the same rate due to gravity, but now with a negative velocity (downwards).
What is the velocity of a projectile when it hits the ground, compared to its launch velocity?	It has the same speed as when it was launched, but in the opposite direction (downwards), because the acceleration due to gravity is constant.
List the four equations for uniform acceleration used in vertical projectile motion.	v = u + at s=\frac{u+v}2\times t v^2 = u^2 + 2as s=ut+\frac12at^2
What is the value of acceleration (a) for the upwards part of the motion?	It is always -9.81ms^{-2} (the acceleration due to gravity, negative because it’s downwards).
What is the value of the initial velocity (u) for the upwards part of the motion?	It is always positive (since it’s launched upwards) and is the velocity it was launched at.
What is the value of the final velocity (v) at the maximum height for the upwards part of the motion?	It is always 0ms^{-1}, because the projectile has been fully decelerated by gravity and is stationary for an instant.
What does displacement (s) represent in the upwards part of the motion?	It is the maximum height reached by the projectile (compared to the launch point), and is positive as the projectile is going up.
What does time (t) represent in the upwards part of the motion?	It is the time taken to reach the maximum height.
What is the value of acceleration (a) for the downwards part of the motion?	It is still -9.81ms^{-2} (the acceleration due to gravity, negative because it’s downwards).
What is the value of the initial velocity (u) at the maximum height for the downwards part of the motion?	It is always 0ms^{-1}, because the projectile is stationary for an instant at the top.
What is the value of the final velocity (v) when the projectile hits the ground?	It is always negative (since it’s falling downwards), and has the same magnitude as the initial launch velocity but negative.
What does displacement (s) represent in the downwards part of the motion?	It is the maximum height reached by the projectile (compared to the launch point), but negative, since the projectile is falling down.
What does time (t) represent in the downwards part of the motion?	It is the time taken to fall back down to the ground.