SUVAT equations question 14

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Christian Lawson-Perfect	said	Ready to use	8 years, 8 months ago
Amy Chadwick	said	Needs to be tested	8 years, 9 months ago

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Christian Lawson-Perfect 8 years, 8 months ago

Gave some feedback: Ready to use

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Christian Lawson-Perfect 8 years, 8 months ago

Gave some feedback: Has some problems

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Amy Chadwick 8 years, 9 months ago

Gave some feedback: Needs to be tested

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Amy Chadwick 8 years, 9 months ago

Created this.

Name	Status	Author	Last Modified
SUVAT equations question 14	Ready to use	Amy Chadwick	04/08/2016 14:16
Simon's copy of SUVAT equations question 14	draft	Simon Thomas	04/04/2019 14:27

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Question statement

A ball is thrown vertically upwards from a point $X$ . The greatest height reached by the ball is $\var{s} \mathrm{m}$ above $X$ . The acceleration due to gravity is $g = 9.8 \mathrm{ms^{-2}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

s	integer	71
a	number	-9.8
s2	integer	2
distance	integer	54
u	number	37.304
t1	number	5.669
t2	number	1.944

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

greatest height reached by the ball

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

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

distance
u

Statement
Advice
"Part a)" - Minimum accepted value
"Part a)" - Maximum accepted value
"Part b)" - prompt

Advice

a)

We are told that the ball is thrown upwards, so let the upwards direction be positive and gravity will act in the opposite direction, thus $a=-9.8$ . At the ball's greatest height, $s=\var{s}$ , the velocity will be $v=0$ for an instant as the ball changes direction and falls downwards. We are trying to find the initial velocity of the ball, $u$ . We have everything except $t$ so we can use the formula $v^2 = u^2 + 2as$ , rearranged for $u$ .

$\begin{align} v^2 & = u^2 + 2as, \\ 0 & = u^2 + \left(2 \times \var{a} \times \var{s}\right), \\ 0 & = u^2 - \var{2*-a*s}. \end{align}$

Therefore

$\begin{align} u^2 &= \var{2*-a*s}, \\ u & = \var{precround ((2*-a*s)^(1/2),3)} \mathrm{ms^{-1}}. \end{align}$

The initial velocity of the ball is $\var{precround ((2*-a*s)^(1/2),3)} \mathrm{ms^{-1}}.$

b)

We have now that $s=\var{distance}$ , $u = \var{u}$ and $a=\var{a}$ . To find the time the ball is above $\var{distance} \mathrm{m}$ we need to find the time the ball first passes over $\var{distance} \mathrm{m}$ and the time it then falls back down past $\var{distance} \mathrm{m}$ , and find the difference between those two times. To find these times we can use the formula $s=ut+\frac{1}{2}at^2$ , rearranged for $t$ .

We have

$\begin{align} s & = ut + \frac{1}{2}at^2, \\ \var{distance} & = \var{u}t - \left( \frac{1}{2} \times \var{-a} \times t^2 \right), \\ 4.9t^2 - \var{u}t + \var{distance} & = 0. \end{align}$

Therefore our first $t$ value, when the ball first passes $\var{distance} \mathrm{m}$ , is

$\begin{align} t & = \frac{\var{u} + \sqrt{\var{u}^2 - \left(4 \times 4.9 \times \var{distance} \right)} }{2 \times 4.9}, \\ & = \var{t1} \mathrm{s}, \end{align}$

and our second $t$ value, when the ball second passes $\var{distance} \mathrm{m}$ , is

$\begin{align} t & = \frac{\var{u} - \sqrt{\var{u}^2 - \left(4 \times 4.9 \times \var{distance} \right)} }{2 \times 4.9}, \\ & = \var{t2} \mathrm{s}. \end{align}$

Then the total time for which the ball is $\var{distance} \mathrm{m}$ or more above $X$ is $\var{t1} - \var{t2} = \var{t1-t2}$ seconds.

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Taxonomy: mathcentre

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Christian Lawson-Perfect 8 years, 8 months ago

Christian Lawson-Perfect 8 years, 8 months ago

Amy Chadwick 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Number entry

Advanced settings

a)

b)

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SUVAT equations question 14

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 8 years, 8 months ago

Christian Lawson-Perfect 8 years, 8 months ago

Amy Chadwick 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Number entry

Advanced settings

a)

b)

Generated value: `integer`