John's copy of SUVAT equations question 13

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John Bridges

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John Bridges 8 years, 6 months ago

Created this as a copy of SUVAT equations question 13.

Name	Status	Author	Last Modified
SUVAT equations question 13	Ready to use	Amy Chadwick	03/08/2016 18:05
Q3 Constant Acceleration	draft	Stephen Bowlzer	10/08/2016 11:11
Q4 Constant Acceleration	draft	Stephen Bowlzer	10/08/2016 12:37
John's copy of SUVAT equations question 13	draft	John Bridges	12/11/2018 13:40
Simon's copy of SUVAT equations question 13	draft	Simon Thomas	04/04/2019 14:17

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Question statement

A ball is thrown vertically upwards with a speed $\var{u} \mathrm{ms^{-1}}$ from a rooftop which is $\var{s} \mathrm{m}$ above ground. The acceleration due to gravity is $g = 9.8 \mathrm{ms^{-2}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	number	-9.8
s	integer	5
u	number	16
t	number	3.5525395635
t2	number	3.553
max_height	number	18.0612244898

Automatically regenerate variables?

Name

Data type

Value

(a number)

Description

Acceleration due to gravity with upwards as positive direction

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `number`

-9.8

→ Used by:

max_height

Advice

Advice

a)

You can model the ball as a particle moving in a straight line with constant acceleration of magnitude $9.8 \mathrm{ms}^{-2}$ due to gravity. As the ball is thrown upwards, we take the positive direction as pointing upwards.

We are told $u=\var{u}$ and $a=\var{a}$ and we are asked to find the greatest height, $s$ , reached by the ball. At its greatest height, the ball will for an instant have a velocity of $v=0$ before it starts to fall back down.

Therefore we can use the equation $v^2=u^2+2as$ , rearranged for $s$ .

Note that the rooftop is $\var{s} \mathrm{m}$ above ground level, so the distance travelled by the ball to its greatest height is $(\simplify{s-{s}}) \mathrm{m}$ .

$\begin{align} v^2 & =u^2+2a(s-\var{s}), \\ 2a(s-\var{s}) & = v^2-u^2, \\ s-\var{s} & = \frac{v^2-u^2}{2a}, \\ & = \frac{0 - \var{u}^2}{2 \times \var{a}}, \\ & = \var{precround(-u^2/(2*a),3)}, \\ s &= \simplify[]{{s}+{precround(-u^2/(2*a),3)}}, \\ &= \var{precround(max_height,3)} \mathrm{m}. \end{align}$

The greatest height reached by the ball is $\var{precround(max_height,3)} \mathrm{m}$ .

b)

The flight time of the ball is the length of time between it being thrown from the rooftop and when it hits the ground below.

Here, the rooftop is $\var{s}\mathrm{m}$ above the ground, therefore the ball will stop moving when it hits the ground at a height $\var{-s} \mathrm{m}$ from its original position.

Therefore we can use the equation $s=ut+\frac{1}{2}at^2$ rearranged for $t$ .

$\begin{align} s & = ut + \frac{1}{2}at^2, \\ -\var{s} & = \var{u}t - \frac{49}{10}t^2, \\ \frac{49}{10}t^2 - \var{u}t - \var{s} &= 0. \end{align}$

Using the quadratic formula, taking only the positive term as we are modelling time, this gives

$\begin{align} t & = \frac{u + \sqrt{u^2 +19.6s)}}{9.8}, \\ & = \frac{\var{u} + \sqrt{\var{u^2} + \left(19.6 \times \var{s}\right)}}{9.8}, \\ & = \var{precround((u+ (u^2 + 19.6*s)^(1/2))/9.8,3)} \mathrm{s}. \end{align}$

The flight time of the ball is $\var{precround((u+ (u^2 + 19.6*s)^(1/2))/9.8,3)}$ seconds.

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This question is used in the following exam:

FS0012 Mechanics Practice CLE by John Bridges in Additional Mechanics CLE Questions.

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John Bridges 8 years, 6 months ago

Error in variable testing condition

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Generated value: `number`

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John's copy of SUVAT equations question 13

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

John Bridges 8 years, 6 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Number entry

Advanced settings

a)

b)

Generated value: `number`