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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Keki on June 12, 2011, 03:35:00 pm

Title: A few questions
Post by: Keki on June 12, 2011, 03:35:00 pm: I've just got a few questions that need answering :3

A stone is thrown horizontally out from the top of the cliff with a velocity of 20 m/s. It lands on the ground beneath after 4.0s.
a) Find the height of the cliff.
80m

b) Find the horizontal travelled by the stone.
80m

c) What is the velocity of the stone just prior to hitting the ground?
44.72 at 63.43° to the horizontal

d) Find the velocity of the stone when it is 1m below the cliff
20.5 m/s at 12.6° to the horizontal

An object is projected with a velocity of 100 m/s at 40° to the horizontal. Find:
a) The horizontal displacement travelled
984.77m

b) The highest point reached from the ground
206.8

c) The velocity at the highest point
76.6 m/s horizontally

d) The speed when it 10m above ground level on the way up
99 m/s

e) The velocity when it is 2m below the highest point on the way down
76.86 m/s to the 4.72° horizontal.

I can't answer the questions in red >_>

Thanks!
Title: Re: A few questions
Post by: DisaFear on June 12, 2011, 03:41:13 pm: In the second question, I used v=u+at to work out the time -
v=0
u= 100 sin 40
a=-9.8
t=?

I got time as 6.56s (this is for half the journey right)

Range (horz. disp) = horizontal velo. x time
= 100 cos 40 x 13.1
=1004.9m

Can anyone confirm?
Title: Re: A few questions
Post by: Vincezor on June 12, 2011, 04:06:35 pm: Quote from: Keki on June 12, 2011, 03:35:00 pm
I've just got a few questions that need answering :3

A stone is thrown horizontally out from the top of the cliff with a velocity of 20 m/s. It lands on the ground beneath after 4.0s.
a) Find the height of the cliff.
b) Find the horizontal travelled by the stone.
c) What is the velocity of the stone just prior to hitting the ground?
d) Find the velocity of the stone when it is 1m below the cliff

a) = 80m
b) = 80m
c) = 44.72m/s
d)
Horizontally:
$u=v=20m/s$

Vertically:
$s=-1, a=-10,u=0 v=?$
$v^{2}=u^{2}+2as$
$v =\sqrt{20}m/s$

Now use pythag to find the velocity

$V = \sqrt{20+ 20^{2}}$
$V \approx 20.493m/s$

EDIT: Forgot the square root something, sorry!
Title: Re: A few questions
Post by: Keki on June 12, 2011, 04:08:34 pm: Just added the answers in now.
Questions are from Jack's notes and they only have answers. No solutions >_>"
Title: Re: A few questions
Post by: Vincezor on June 12, 2011, 04:30:53 pm: Quote from: Keki on June 12, 2011, 03:35:00 pm

An object is projected with a velocity of 100 m/s at 40° to the horizontal. Find:
a) The horizontal displacement travelled
984.77m

b) The highest point reached from the ground
206.8

c) The velocity at the highest point
76.6 m/s horizontally

d) The speed when it 10m above ground level on the way up
99 m/s

e) The velocity when it is 2m below the highest point on the way down
76.86 m/s to the 4.72° horizontal.

a)
Vertically:
$u = 100sin40 v=0 a=-10 t=?$

$t=\frac{v-u}{a}$

$t\approx 6.43s$

as time up = time down

$Time total = 12.844s$

Horizontally:
$s=u*t$
s=100cos40*12.86
$s\approx 984.81m$

d)
Vertically
$s=10, u=100sin40, v=?, a=-10$

$v^{2}=u^{2}+2as$
$v= 65.82m/s<br />$
Use Pythag once more

$V =\sqrt{65.82^2+(100cos40)^{2}}$
$V \approx 100.995m/s$

e)
Vertically
$s=-2, u=0, v=?, a=-10$

$v^{2}=u^{2}+2as$
$v\approx 6.32m/s$

Use Pythag once more

$V =\sqrt{6.32^2+(100cos40)^{2}}$
$V \approx 76.865m/s$

I see you're having trouble in the same areas. My advice is to break the question up into it's horizontal and vertical components. Write down what you are given and what you need to find. Then use the SUVAT questions to find the answer.
Title: Re: A few questions
Post by: DisaFear on June 12, 2011, 04:33:40 pm: I see that using 9.8 as gravity made a massive difference :o