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# Projectile Motion and What if Velocity and Acceleration are in Different Directions?

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## Projectile Motion and What if Velocity and Acceleration are in Different Directions?

Vectors at right angles to each other are independent.

If you are considering the effect of two (or more) vectors on an object, it is important to remember that:

**Vectors at right angles to each other do not have any effect on each other.**

An easy example; no matter how hard you push down on an object you will never make it accelerate sideways.

**Projectile Motion - ignoring friction**

If a stone is thrown horizontally from a cliff top it follows what is called **projectile motion**.

**Vertically** it has constant acceleration downwards (due to gravity).

**Horizontally** it has constant velocity (for instance, no acceleration or deceleration as there is no friction).

Here's where vector components become important. (See **Vectors and Scalars Learn-it**)

Whatever happens vertically has no effect horizontally. So use equations of motion to find what's going on but only consider either horizontally or vertically at any one time.

**Worked example:**

**A stereo is thrown off a cliff 100 m high at 10ms ^{-1} whilst playing Cliff Richard songs at full volume. How far does it travel?**

**Horizontally**: When t = 0, u = 10 ms^{-1}, a = 0

**Vertically:** When t = 0, u = 0, a =10ms^{-2} (** Note:** I've chosen down to be positive)

Use the equations of motion with **either** the **horizontal** or **vertical** information, but **never mix** information from both.

The only thing that is going to be exactly the same both horizontally and vertically at any moment is **time, t.**

**If we look vertically first:**

List know quantities first:

u = 0 m/s

s = +100 m

a = +10 m/s^{2}

time to reach ground below cliff = t seconds

So, use equation s = ut + ½ at²

100 = 0.t + ½ at²

100 + 5t²

t² = 20

t = √ 20 = 4.47s

**It travels horizontally for this same time,**

So, u = 10 m/s

t = 4.47s

a = 0 m/s²

s = horizontal distance from landing to the base of cliff

s = ut + ½ at²

s = 10. 4.47 + ½. 0 . (4.47)² = 44.7 + 10

= 44.7 m

So the stereo will travel a distance of 44.7 m away from the cliff before it reaches it stops playing Cliff Richard.

**Cannon Balls - a tricky but common example (ignore friction)**

Consider a cannon ball fired at an angle of 30 degrees above the horizontal. Once again, think about the **horizontal** and **vertical** components of the motion one at a time. This time you have to use trigonometry to find the components of u in the two directions.

*Initially:*

**Horizontally:** At t = 0s, u = 10 cos30 m/s, a = 0 m/s²

**Vertically:** At t = 0, u = 10 sin30 m/s, a = -10 m/s² (** Note:** the acceleration and initial vertical velocity are in different directions - velocity is up (positive) and acceleration is down (negative)

*Some Useful Tricks:*

To find out about the ball at the highest point in its flight, remember that at that point **vertically:** v = 0 m/s.

(For that instant it is travelling horizontally so it has no vertical velocity at all).

To find the time for the whole flight you usually have to find the time for **half the flight** by considering the time for the vertical velocity to reduce to zero from its initial value (for instance, the time it takes the ball to stop moving any higher) and then double it.

*Question:*

**Find the range of a golf ball that is hit and leaves the ground at 20 m/s at an angle of 25 degrees to the horizontal.**

**Calculate the vertical and horizontal parts of u, v and a, by dragging the correct values into the equations:**

**Using the information above, fill in the correct answers in the boxes below:**

**Direction is important**

To show different directions we use a positive or negative sign. It doesn't matter whether you choose up or down, left or right as positive, as long as you stick to it for the rest of the question.

*For Example:*

If you are going to the right at 10ms^{-1} but accelerating to the left (for instance, decelerating) at 2ms^{-2} then u = +10ms^{-1} and a = -2ms^{-2} i.e. + means to the right and - means to the left.

** Remember:** you can only work in one dimension at a time!

**Acceleration due to gravity**

Objects in a gravitational field experience a downward force, their weight. If unbalanced, this will produce a downward acceleration. This crops up frequently in A-level questions. However, it's easy to deal with. Simply always use acceleration as:

a = g = 9.81 ms^{-2} downwards.

**For Example:**

Drop a stone from a cliff.

Initially, t = 0, u = 0, and a = + 9.81 ms^{-2} (* Note*: I've chosen

**down**to be positive here)

**Or**

Throw a stone upwards at 10 ms^{-1}

Initially, t = 0, u = 10ms^{-1} and a = g = -9.81ms^{-2}. (* Note*: I've chosen

**up**to be positive here to show that it doesn't matter which one you choose as long as you're consistent.)

**Question:**

**A skydiver jumps from a height of 980 m. Gravity is 10 m/s ^{2}.**