Question:

The point A is 1.8 m vertically above horizontal ground.

At time t = 0, a small stone is projected vertically upwards with speed U m s−1

from the

point A.

At time t = T seconds, the stone hits the ground.

The speed of the stone as it hits the ground is 10 m s−1

In an initial model of the motion of the stone as it moves from A to where it hits

the ground

• the stone is modelled as a particle moving freely under gravity

• the acceleration due to gravity is modelled as having magnitude 10 m s−2

Using the model,

(a) find the value of U

suvat answer:

s = -1.8

u = U

v = -10

a = -10

t = ?

I understand why you use these values and that upwards is the positive direction but I dont get why the displacement is negative and the final velocity is negative. Is the answer based on the whole upwards and downwards motion of the stone or is it based on the upwards/downwards motion of the stone only?

The point A is 1.8 m vertically above horizontal ground.

At time t = 0, a small stone is projected vertically upwards with speed U m s−1

from the

point A.

At time t = T seconds, the stone hits the ground.

The speed of the stone as it hits the ground is 10 m s−1

In an initial model of the motion of the stone as it moves from A to where it hits

the ground

• the stone is modelled as a particle moving freely under gravity

• the acceleration due to gravity is modelled as having magnitude 10 m s−2

Using the model,

(a) find the value of U

suvat answer:

s = -1.8

u = U

v = -10

a = -10

t = ?

I understand why you use these values and that upwards is the positive direction but I dont get why the displacement is negative and the final velocity is negative. Is the answer based on the whole upwards and downwards motion of the stone or is it based on the upwards/downwards motion of the stone only?

Original post by ljaiju

Question:

The point A is 1.8 m vertically above horizontal ground.

At time t = 0, a small stone is projected vertically upwards with speed U m s−1

from the

point A.

At time t = T seconds, the stone hits the ground.

The speed of the stone as it hits the ground is 10 m s−1

In an initial model of the motion of the stone as it moves from A to where it hits

the ground

• the stone is modelled as a particle moving freely under gravity

• the acceleration due to gravity is modelled as having magnitude 10 m s−2

Using the model,

(a) find the value of U

suvat answer:

s = -1.8

u = U

v = -10

a = -10

t = ?

I understand why you use these values and that upwards is the positive direction but I dont get why the displacement is negative and the final velocity is negative. Is the answer based on the whole upwards and downwards motion of the stone or is it based on the upwards/downwards motion of the stone only?

The point A is 1.8 m vertically above horizontal ground.

At time t = 0, a small stone is projected vertically upwards with speed U m s−1

from the

point A.

At time t = T seconds, the stone hits the ground.

The speed of the stone as it hits the ground is 10 m s−1

In an initial model of the motion of the stone as it moves from A to where it hits

the ground

• the stone is modelled as a particle moving freely under gravity

• the acceleration due to gravity is modelled as having magnitude 10 m s−2

Using the model,

(a) find the value of U

suvat answer:

s = -1.8

u = U

v = -10

a = -10

t = ?

I understand why you use these values and that upwards is the positive direction but I dont get why the displacement is negative and the final velocity is negative. Is the answer based on the whole upwards and downwards motion of the stone or is it based on the upwards/downwards motion of the stone only?

At A, displacement 0 occurs at t=0, so the final displacement is 1.8m below that position so s=-1.8 as its negative downwards. The final velocity is also heading downwards so v=-10.

It may help to sketch the parabola of vertical displacement against time. If youre unsure, post what you think.

(edited 1 month ago)

Original post by mqb2766

displacement 0 occurs at t=0, so the final displacement is 1.8m below that position so s=-1.8 as its negative downwards. The final velocity is also heading downwards so v=-10.

It may help to sketch the parabola of vertical displacement against time. If youre unsure, post what you think.

It may help to sketch the parabola of vertical displacement against time. If youre unsure, post what you think.

that is fine, i understand it much better now thank youu.

- I don't like suvat equations, can I sketch graphs instead?
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