Edmund's Physics Questions

[Edmund]

Can someone help me with this... I forgot how to do this...

A ball is thrown vertically upward with a speed of 19.6 m/s. What is the ball's velocity and its height after 1.0, 2.0, 3.0 and 4.0s?

[TrueTears]

u = -19.6 a = 9.8 t = 1

then just plug it in v = u + at to work out velocity

and then plug it in for displacement above ground

(let up be -ve and down be +ve)

same goes for t = 2 3 and 4

[Flaming_Arrow]

use u=19.6 v=? and a=-10

[Edmund]

Boo... Calculation mistake lolol

Thanks

[MattPritchard]

u = -19.6 a = 9.8 t = 1

then just plug it in v = u + at to work out velocity

and then plug it in for displacement above ground

same goes for t = 2 3 and 4

To avoid confusion I think that it would actually be better to make u (initial velocity) positive (19.6m/s) and a (acceleration) negative (-9.8m/s), to show that it is decellerating and when the velocity becomes negative it is moving in the other direction (which would be downwards).

As stated before
-To find Velocity

u=19.6m/s Up (Initial Velocity)
v=???m/s Up or Down (Final Velocity)
a=-9.8m/s (Acceleration)
t=t seconds (Time) *Substitute t values 1.0,2.0,3.0 etc

-To find Displacement

u=19.6m/s Up (Initial Velocity)
x=???m Up or Down (Displacement from origin)
a=-9.8m/s (Acceleration)
t=t seconds (Time) *Substitute t values 1.0,2.0,3.0 etc

[TrueTears]

u = -19.6 a = 9.8 t = 1

then just plug it in v = u + at to work out velocity

and then plug it in for displacement above ground

same goes for t = 2 3 and 4

To avoid confusion I think that it would actually be better to make u (initial velocity) positive (19.6m/s) and a (acceleration) negative (-9.8m/s), to show that it is decellerating and when the velocity becomes negative it is moving in the other direction (which would be downwards).

As stated before
-To find Velocity

u=19.6m/s Up (Initial Velocity)
v=???m/s Up or Down (Final Velocity)
a=-9.8m/s (Acceleration)
t=t seconds (Time) *Substitute t values 1.0,2.0,3.0 etc

-To find Displacement

u=19.6m/s Up (Initial Velocity)
x=???m Up or Down (Displacement from origin)
a=-9.8m/s (Acceleration)
t=t seconds (Time) *Substitute t values 1.0,2.0,3.0 etc

Yeap you could do that, works out the same

[MattPritchard]

Yeap you could do that, works out the same

Thought so, but in the exam do they care whether it is negative or positive providing that you give the direction?

-5m/s east = 5m/s west

[Flaming_Arrow]

Yeap you could do that, works out the same

Thought so, but in the exam do they care whether it is negative or positive providing that you give the direction?

-5m/s east = 5m/s west

i think it'd be better to write the positive direction

[MattPritchard]

Yeap you could do that, works out the same

Thought so, but in the exam do they care whether it is negative or positive providing that you give the direction?

-5m/s east = 5m/s west

i think it'd be better to write the positive direction

Yeah i thought so, you would just put that in working out wouldnt you when you got a negative answer?

[TrueTears]

Yeap you could do that, works out the same

Thought so, but in the exam do they care whether it is negative or positive providing that you give the direction?

-5m/s east = 5m/s west

i think it'd be better to write the positive direction

Yeah i thought so, you would just put that in working out wouldnt you when you got a negative answer?

that's right

[Edmund]

Another question, this time 2D kinematics:

A cannon is fired from the top of a castle wall at a speed of 50m/s and an angle of 30 degrees. The height of the wall is 11.025m.

How far from the castle wall does the cannonball hit the ground?

[TrueTears]

HINT: split everything into its vertical and horizontal components

[Edmund]

x comp
u = 50cos30
t = ?
x = ?

y comp
Find time here.

Problem is, Im not getting the correct time

[Mao]

1. find time it takes to get to the top. (verticle velocity = 0)
2. find distance travelled from point of launch to top.
3. add above to the height of wall, that is the distance it travels on the way down
4. find time it takes to crash into ground
5. add [1] and [4] together
6. find range
7. ??
8. PROFIT!

[kamil9876]

OR:

Y component:

Assume upwards is positive, hence the particle must have a displacement of -11.025

s=0.5at^2+ut

s=-11.025
a=-g
t=thing u wanna find out
u=vertical component of initial velocity.