Motion in a vertical Plane
This chapter covers motion in a vertical place. It covers applying the constant acceleration formulae to solve problems relating to vertical motion under gravity and the constant acceleration formulae.
What is gravity?
Gravity is the force that planets exert on objects. In 1660 Sir Isaac Newton produced a universal law of gravitation which produced a formula for the effect of gravity anywhere in the universe. Isaac was inspired to produce the formula after observing an apple falling from a tree.
On Earth the acceleration of a falling object due to gravity is 9.8ms-2 if the effect of air resistance is ignored. Gravitation depends on the mass of an object that means on another planet the gravity may be less or more than 9.8ms-2. Force is measured in Newtons (N). The weight on an apple is approximately 1N.
Dropping an object
The acceleration due to gravity is constant. That means we can use the constant acceleration formulae to work out how a falling object behaves under the pull of gravity.
Example
A stone is dropped from a cliff and lands in the sea 4 seconds later. How high is the cliff?
It is best to draw a labelled diagram indicating the directions taken. The diagram is shown below;
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We must relate the information that we have to; S, u, v, a, t and select a formulae connecting S, u, a, and t. The formula is;
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S stands for distance. We substitute the known values in the formula to get;
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We say that the cliff is 78.4m high.
Example
A coin is dropped into a well. The depth of the well is 23m. How long does it take for the coin to reach the water?
First we draw a simple diagram to show the known information and what we’re trying to find.
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We can relate the known information to S, u, v, a, t. We need to select a constant acceleration formula connecting S, u, a, and t. We substitute the known values in the formula.
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We make t the subject to get;
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We can conclude that the time taken is 2.2 seconds.
Projecting an object downwards
When projecting an object downwards you have to consider that its initial velocity isn’t zero. Here is an example.
A stone is thrown downwards with a speed of 4ms-1 and hits the ground with a speed of 18ms-1. Calculate the height from which the stone was thrown. First we draw a diagram for the problem as shown below.
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Here we can use the formula;
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We then substitute in the known values.
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We change to make h the subject;
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The height from which the stone is thrown is;
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Projecting an object upwards
In the following example the object is being project upwards.
Example
An apple is thrown vertically upwards with speed 8ms-1 from a height of 1.6m above the ground.
- Find the greatest height reached by an apple
- Calculate the time taken for the apple to reach the ground
First we draw a diagram to get a graphical representational of what is happening.
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Above we’ve noted that v=0. This is because at maximum height the velocity is 0.
We can use the following formula;
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If we substitute in the known values we get;
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We make h the subject;
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The greatest height reached is 3.27m above the point of projection.
We can now add the height of project to find the greatest height reached above the ground. That is;
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The greatest height reached is 4.87 above the ground.
We know that the displacement when it hits the ground is negative as it is 1.6m below the start. So we have;
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We use the formula;
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We substitute in the known values;
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To find t we must use the quadratic formula.
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Time cannot be negative so the ball hits the ground after 1.8 seconds.
