projectile motion questions =( (1 Viewer)

[coeyz]

1) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s

a) Calculate their time to fall to the bottom of the cliff. (4.52s)
b) Calculate their velocity half way down. (42.3 m/s)
c) Calculate the time their speed will be 30.0 m/s (3.06s)
d) Calculate the speed at which they hit the ground (62.613 m/s)
--------------------

2) Projectile is fired at 30m/s and has a range of 90m. It is in flight for 6s.

a) the initial velocity of the projectile
(30m/s at 60* to horizontal)



THANKS A LOT!!!

 

[random-1005]

1) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s

a) Calculate their time to fall to the bottom of the cliff. (4.52s)
b) Calculate their velocity half way down. (42.3 m/s)
c) Calculate the time their speed will be 30.0 m/s (3.06s)
d) Calculate the speed at which they hit the ground (62.613 m/s)
--------------------

2) Projectile is fired at 30m/s and has a range of 90m. It is in flight for 6s.

a) the initial velocity of the projectile
(30m/s at 60* to horizontal)



THANKS A LOT!!!

a.
delta (d) y = 200
ux= 0.6m/s
ay (acceleration)=9.8ms squared
uy=0
take downwards as positive
s=uyt +1/2ayt^2
200= 0 +4.9t^2
t^2=200/4.9
t= 6.38 sec

b.
we want velocity when delta y =100
ie we want half of the trip time
ie t= 3.19

v=u+at
down is positive
v=o +9.8(3.19)
v= 31.26m/s downwards

c. time when v = 30m/s
v=u+at
30=0+9.8t
t=30/9.8
t=3.06 seconds

d. speed at which it hits the ground
ie speed when delta y (d)=200

v^2=u^2+2as
v^2=0+2(9.8)(200)
v^2=3920
v=62.61m/s down

2.
delta x (x) = 90m
trip time = 6 seconds
x=ux multiplied by trip time
therefore ux= 90/6
ux=15m/s


drawing a vector diagram
and using trigonmetry
cosx = 15/30
cosx=1/2
therefore x= 60 degress therefore velocity=30m/s at 60 degress to horizontal


u did ok, must have made a calculation error in the first part of q1 , and it followed through for the rest of part 1, thats why they all look wrong when it was most likely a simple calculator mistake.

 

[shady145]

1.
a)
200=9.8xt^2
t^2=20.41
t=4.52 seconds
b)
v^2=u^2+2ay
v^2=0^2+19.6x100
v^2=1960
v=44.27 in the y component
use vectors to find the velocity at this instant
v=root(.6^2+44.27^2)
v= the rwong answer (44.27 again) and an angle is needed if u asked for velocity
c)u worded this q wrong, i think you mean to say how long until the velocity reaches 30m/s
d)
v=u+at
v=0+9.8x4.52
v=44.3m/s
vectors to find final speed
44.3^2+.6^2=final speed squared
final speed= 44.3, which is wrong according to ur answers.

2.
a)
find the vectors to find the angle
initial velocity is obviously 30m/s
s(x)=u(x)xtime
90=u(x)x6
u(x)=15m/s
u(x)=vcosangle
cosangle=15/30
angle=60
therefore final velocity = 30m/s at 60* to horizontal

 

[shady145]

omg, making the same mistake myself, thats what u get for trying to use the formulas of the top of ya head lol.

 

[random-1005]

1.
a)
200=9.8xt^2
t^2=20.41
t=4.52 seconds
b)
v^2=u^2+2ay
v^2=0^2+19.6x100
v^2=1960
v=44.27 in the y component
use vectors to find the velocity at this instant
v=root(.6^2+44.27^2)
v= the rwong answer (44.27 again) and an angle is needed if u asked for velocity
c)u worded this q wrong, i think you mean to say how long until the velocity reaches 30m/s
d)
v=u+at
v=0+9.8x4.52
v=44.3m/s
vectors to find final speed
44.3^2+.6^2=final speed squared
final speed= 44.3, which is wrong according to ur answers.

2.
a)
find the vectors to find the angle
initial velocity is obviously 30m/s
s(x)=u(x)xtime
90=u(x)x6
u(x)=15m/s
u(x)=vcosangle
cosangle=15/30
angle=60
therefore final velocity = 30m/s at 60* to horizontal

part 1 d is wrong because u did Q1 a wrong, the correct equation is s=ut+1/2at^2 for question 1 a, u used s=ut+at^2 u omitted the half in front of at^2
there arent hard to remember

v+u+at
V^2+u^2+2as
s=ut+1/2at^2

 

[random-1005]

hey shady, u dont need an angle of projection because the object is falling vertically down, the angle is effectively zero

 

[shady145]

hehe u do need an angle for a velocity as u need both magnitude and direction in your answer to get the mark

 

[shady145]

random ur part c is wrong. the q says Calculate the time their speed will be 30.0 m/s. the answer is an infinite small amount of time. but i think he worded the question wrong in which case ur answer would be the right if he adjusted q.

 

[darkchild69]

1 a)

Ok, firstly what do we know?

a = -9.8ms^-2
Sy = -200m
uy = 0
t = ?

s = ut +1/2at^2

-200 = 0 + -4.9t^2

t^2 = -200/-4.9

t = 6.39s

The answer given (4.52s) is incorrect

b)

So, at halfway down Sy = -100

It is not sufficient just to take half of the time as random did as the object is accelerating, it will cover the second 100m in a time less than it will cover the first 100m!

So

using the same equation

-100 = 0 + -4.9t^2

t^2 = -100/-4.9

t = 4.52s

So it has taken the lemming 4.52s to fall the first 100m

you then need to figure out the velocity after this amount of time:

v = u + at

v = 0 + -9.8*4.52

v = 44.27m/s

now that you have the vertical component of velocity and the question says the lemming falls off at a velocity of 0.6ms-1, assuming this is the horizontal component of velocity and as Vx = Ux

then:
Vx = 0.6
Vy = 44.27


Tan theta = 0.6/44.27

theta = Tan^-1(0.6/44.27)

theta = 0.78 degrees

v^2 = (Vy^2 + Vx^2)^1/2
v^2 = (44.27)^2 + (0.6)^2)^1/2

v = 44.28ms-1 at an angle of 89.22 degrees below the horizontal



c)

Vy = -30
Uy = 0
ay = -9.8
t = ?

v = u + at

t = v-u/a

t = -30/-9.8

t = 3.06s


d) time of flight is 6.39s

therefore:

Vy = ?
Uy = 0
ay = -9.8
t = 6.39

v = u + at

v = 0 + 9.8*6.39

v = 62.622ms-1



2a)
t = 6s
Sx = 90m

therefore Ux = 15ms-1

Ux = U Cos theta

15 = 30 Cos theta

Cos theta = 15/30

Cos theta = 0.5

theta = Cos^-1(0.5)

theta = 60 degrees

therefore initial velocity is 30ms-1 60 degrees above horizontal


FINALLY GOT TO A QUESTION BEFORE PWNAGE101! w0000t!

 

[shady145]

thats because it had already been answered lol

 

[darkchild69]

thats because it had already been answered lol

But no one had gotten all the answers correct!

 

[random-1005]

1 a)

Ok, firstly what do we know?

a = -9.8ms^-2
Sy = -200m
uy = 0
t = ?

s = ut +1/2at^2

-200 = 0 + -4.9t^2

t^2 = -200/-4.9

t = 6.39s

The answer given (4.52s) is incorrect

b)

So, at halfway down Sy = -100

It is not sufficient just to take half of the time as random did as the object is accelerating, it will cover the second 100m in a time less than it will cover the first 100m!

So

using the same equation

-100 = 0 + -4.9t^2

t^2 = -100/-4.9

t = 4.52s

So it has taken the lemming 4.52s to fall the first 100m

you then need to figure out the velocity after this amount of time:

v = u + at

v = 0 + -9.8*4.52

v = 44.27m/s

now that you have the vertical component of velocity and the question says the lemming falls off at a velocity of 0.6ms-1, assuming this is the horizontal component of velocity and as Vx = Ux

then:
Vx = 0.6
Vy = 44.27


Tan theta = 0.6/44.27

theta = Tan^-1(0.6/44.27)

theta = 0.78 degrees

v^2 = (Vy^2 + Vx^2)^1/2
v^2 = (44.27)^2 + (0.6)^2)^1/2

v = 44.28ms-1 at an angle of 89.22 degrees below the horizontal



c)

Vy = -30
Uy = 0
ay = -9.8
t = ?

v = u + at

t = v-u/a

t = -30/-9.8

t = 3.06s


d) time of flight is 6.39s

therefore:

Vy = ?
Uy = 0
ay = -9.8
t = 6.39

v = u + at

v = 0 + 9.8*6.39

v = 62.622ms-1



2a)
t = 6s
Sx = 90m

therefore Ux = 15ms-1

Ux = U Cos theta

15 = 30 Cos theta

Cos theta = 15/30

Cos theta = 0.5

theta = Cos^-1(0.5)

theta = 60 degrees

therefore initial velocity is 30ms-1 60 degrees above horizontal


FINALLY GOT TO A QUESTION BEFORE PWNAGE101! w0000t!

lol thanks for showing my mistakes , its been like 2 months since i did a projectile question, lol

 

[4theHSC]

1) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s

a) Calculate their time to fall to the bottom of the cliff. (4.52s)
b) Calculate their velocity half way down. (42.3 m/s)
c) Calculate the time their speed will be 30.0 m/s (3.06s)
d) Calculate the speed at which they hit the ground (62.613 m/s)
--------------------

2) Projectile is fired at 30m/s and has a range of 90m. It is in flight for 6s.

a) the initial velocity of the projectile
(30m/s at 60* to horizontal)



THANKS A LOT!!!

*sigh* this question has already been answered, get the revised edition next time.

 

[Pwnage101]

FINALLY GOT TO A QUESTION BEFORE PWNAGE101! w0000t!

lol congrats! :rofl:

yeh, havent been on BoS much over the past 4 days, been very busy