Vertical motion under gravity

In a nutshell

Gravity acts downwards on all objects, giving a free-falling object an constant downward acceleration of $$9.8m/s^2$$. For gravity to provide a constant acceleration, air resistance must be ignored. You can use previously derived formulae to solve problems involving gravity. 

Equations

Variable definitions

Gravity

Gravity is constantly acting downwards on objects, giving them an approximate downward acceleration of $$9.8ms^{-2}$$​. Gravity is represented by the letter $$g$$. When dealing with gravity, there is an assumption that there is no air resistance. This results in there being no differences due to the mass of an object - they all accelerate downwards at the same rate.

When dealing with vertical motion, you can take the positive direction to be either upwards or downwards.

Note: Sometimes a question may ask you to use a different value of $$g$$, such as $$10ms^{-2}$$, but unless the question tells you otherwise, always take $$g$$ to be $$9.8ms^{-2}$$.

Example 1

A ball falls off a table. The table-top is $$80cm$$ above the floor. Find:

a: The time it takes the ball to reach the floor.

b: The speed of the ball as it hits the floor. Give your answer to $$3$$​ significant figures.

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First, decide on a positive direction. As the ball only moves downwards, make downwards the positive direction. Next, make a note of all the known variables: $$u=0ms^{-1},\ a=9.8ms^{-2},\ s=0.8m$$.

a: Final velocity is not given, so use a formula without $$v$$:

$$s=ut+\dfrac{1}{2}at^2$$

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Subsitute known variables:

$$0.8=0t+\dfrac{1}{2}(9.8)t^2$$​​

Simplify:

$$0.8=4.9t^2$$​​

Solve for $$t$$:

$$t^2= \dfrac{0.8}{4.9}$$​​

$$t=\sqrt{\dfrac{0.8}{4.9}}=0.404061$$​​

Give the answer to 3 significant figures:

The ball takes $$\underline{ 0.404\ s}$$ to hit the floor.

b: Use a formula which does not include $$t$$:

$$v^2 = u^2 +2as$$

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Substitute known values and solve:

$$v^2 = 0^2 +2(9.8)(0.8)$$​​

$$v^2 = 15.68$$​​

$$v=\sqrt{15.68} = 3.95980$$​​

The speed of the ball as it hits the floor is $$\underline{3.96 \ m.s^{-1}}$$.

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Example 2: 

A ball is kicked upwards from a point $$2m$$ above the ground with a speed of $$15ms^{-1}$$. Work out:

a: The greatest height above the ground the ball reaches. Give your answer to $$3$$​ significant figures.

b: The time taken for the ball to reach the ground again.

Take the positive direction to be upwards. Known variables: $$u=15ms^{-1}, a=-9.8ms^{-2}$$.

a: When the ball reaches its greatest height, it is just before it makes its way back down to the ground. At this moment, it is neither moving up nor down, therefore $$v=0ms^{-1}$$.

To work out displacement without needing time, use: 

$$v^2 = u^2 +2as$$​​

Substitute known values:

$$0^2 = 15^2 + 2(-9.8)s$$​​

Rearrange for $$s$$ and solve:

$$0=225-19.6s$$

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$$s=\dfrac{225}{19.6}= 11.4795918$$​​

The ball started $$2m$$ above the ground, therefore this has to be added to the displacement to get the total height above the ground:

$$Total\ height= s+2 =11.4795918+2=13.4795918$$​​

The greatest height above the ground the ball reaches is $$\underline{13.5 \ m}$$.

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b: When the ball reaches the ground again, $$s=-2$$ because it started $$2m$$ above the ground.

To work out time without the final velocity:

​$$s=ut+\dfrac{1}{2}at^2$$​​

Substitute known variables:

​$$-2=15t+\dfrac{1}{2}(-9.8)t^2$$​​

​$$-2=15t-4.9t^2$$​​

Rearrange to form a quadratic:

​$$4.9t^2-15t-2=0$$​​

Solve using the quadratic formula:

​$$t=\dfrac{-b\pm \sqrt{(b^2-4ac)}}{2a}$$​​

$$t=\dfrac{-(-15)\pm \sqrt{((-15)^2-4(4.9)(-2))}}{2(4.9)}$$​​

​$$t=\dfrac{15\pm \sqrt{225+39.2}}{9.8}$$​​

​$$t=\dfrac{15\pm 16.2542}{9.8}$$​​

​$$t=3.18921, t=-0.127983$$​​

Time can't be negative, therefore take the positive answer and round to $$3$$ significant figures:

The ball takes $$\underline{3.20 \ s}$$ to reach the ground again.

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