Question 3.2.1: Additional Answer Options

2019 - May/June - Paper 1 - Question 3

3.1

Motion of an object under the influence of gravity/gravitational force (weight) only.

Motion in which the only force acting on the object is gravity/weight.

Vertical motion in which friction/air resistance is absent.

Motion in the air with an acceleration of 9,8 m∙s^-2.

The focus of your answer must be on the motion of the object and not on the object.

2 marks

(1 x 2 marks)

3.2.1

Option 1:

Upwards as positive:

$v_{f}=v_{i}+a\Delta t$

$0=v_{i}+(-9,8)(1,53)$

$\therefore\mathrm{v}_{\mathrm{i}}=14,99\mathrm{~m}\cdot\mathrm{s}^{-1}\left(15\mathrm{~m}\cdot\mathrm{s}^{-1}\right)$

Downwards as positive:

$v_{f}=v_{i}+a\Delta t$

$0={v}_{i}+(9,8)(1,53)$

$\therefore\mathrm{v}_{\mathrm{i}}=-14,99\mathrm{~m}\cdot\mathrm{s}^{-1}$

$v_{i}=14,99\mathrm{~m}\cdot\mathrm{s}^{-1}\left(15\mathrm{~m}\cdot\mathrm{s}^{-1}\right)$

Tips:

The velocity of an object which is thrown upwards is 0 m·s^-1 at the highest point in its motion.
Remember, velocity and acceleration are vector quantities, therefore, they have a direction.

There are more ways to answer this question. To view other options, click here.

3 marks

(3 x 1 mark)

3.2.2

How to approach this question:

The most direct method to answer this question uses the initial velocity calculated in question 3.2.1, the time taken to reach the highest point in the motion as well as the acceleration due to gravity.
Remember, the direction for all of the vector quantities.

Option 1:

Upwards positive:

$\Delta y=v_{i}\Delta t+1/2a\Delta t^2$

=14.99(1.53)+1/2(-9.8)(1.53)^2

$=11,47\mathrm{m}(11,46-11,48)$

Maximum height is 11,47 m.

You do not need to add the 40m height of the building onto this displacement since the question specifically asks for the height that the ball reaches above the edge of the roof.

Downwards as positive:

$\Delta\mathrm{y}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2$

=-14,99(1,53)+1/2(9,8)(1,53)^2

$=-11,47\mathrm{~m}(11,46-11,48)$

Maximum height is 11,47 m.

There are more ways to answer this question. To view other options, click here.

3 marks

(3 x 1 mark)

3.3

Option 1:

Upwards positive:

$\Delta\mathrm{y}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2$

=(14,99)(4)+1/2(-9,8)(4)^2

$=-18,4\mathrm{~m}$

The position is 18,4 m downwards.

Downwards as positive:

$\Delta\mathrm{y}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2$

=(-14,99)(4)+1/2(9,8)(4)^2

$=18,4\mathrm{~m}$

The position is 18,4 m downwards.

Tips:

The most direct method to answer this question uses the initial velocity calculated in question 3.2.1, the time taken to reach the highest point in the motion which is 4 seconds as well as the acceleration due to gravity.
Remember, vector quantities require a direction.

There are more ways to answer this question. To view other options, click here.

3 marks

(3 x 1 mark)

3.4

No.

The motion of the ball is only dependent on its initial velocity / the initial
velocity depends on the time taken to reach maximum height.

Tips:

If the building is 10 m shorter, this will not change the answer to question 3.2 because the ball can still be thrown up with the same velocity and reach the same maximum height above the edge of the roof.
The answer to question 3.3 indicates that 4 seconds after being thrown the ball will have fallen 18,4m below the edge of the roof, since the building is taller than 18,4m, this also will not change the answer.

3 marks

(3 x 1 mark)

Option 2

$\mathrm{F}_{\text{net }}=\mathrm{ma}$