how to slove this question (1 Viewer)

[Fply]

During a visit to the Moon an astronaut throuws a rock vertically and it reaches a height of 20 m Acceleration due to gravity on moon is 1.6 what is the speed at which it was thrown on the moon?

 

[ND]

s=20
a=-1.6
v=0
u=?

Use v^2=u^2+2as.

 

[failingTheHsc]

y acceleration = -1.6
y speed = -1.6t + V, V = inital velocity
y displacement = -1.6/2 t^2 + Vt

max heigh occurs when y speed = 0
t = v/1.6

sub back in to displacement
20 = -1.6/2 * V^2/(1.6)^2 + V^2/16
then solve for V
V = root(40 * 1.6)

i doubt they will ask this, it is too 'matemetically complicated'

 

[freaking_out]

Originally posted by failingTheHsc
...i doubt they will ask this, it is too 'matemetically complicated'

lol, true that...they are maths haters.

 

[ND]

Originally posted by failingTheHsc
y acceleration = -1.6
y speed = -1.6t + V, V = inital velocity
y displacement = -1.6/2 t^2 + Vt

max heigh occurs when y speed = 0
t = v/1.6

sub back in to displacement
20 = -1.6/2 * V^2/(1.6)^2 + V^2/16
then solve for V
V = root(40 * 1.6)

i doubt they will ask this, it is too 'matemetically complicated'

Don't make it more complicated than it needs to be, look at my above post.

 

[Fply]

I have the same though as you guys
but if I tell you the answer is 25.3
can you explain why ?

 

[ND]

The answer 25.3 is wrong, simple as that.

 

[dnt stop runnin]

i cant c how the hell they got 25.3....it cant be.....im getting 8 like everyone else..i agree with ND, answer is wrong.

 

[Fply]

this is a multipal choice question comes from James Rues High


the actual answer is 25.3

the other choices of answer include 62.6 640 and 3920

 

[Takuya]

Uh this is a very simple problem...

Using v^2=u^2+2ar

The answer is u=8m/s

 

[Takuya]

Maybe someone copied out the question incorrectly...

 

[Takuya]

Originally posted by failingTheHsc
y acceleration = -1.6
y speed = -1.6t + V, V = inital velocity
y displacement = -1.6/2 t^2 + Vt

max heigh occurs when y speed = 0
t = v/1.6

sub back in to displacement
20 = -1.6/2 * V^2/(1.6)^2 + V^2/16
then solve for V
V = root(40 * 1.6)

i doubt they will ask this, it is too 'matemetically complicated'

You're the one who's making it too complicated by using a whole lot of steps.

 

[failingTheHsc]

Originally posted by Takuya
You're the one who's making it too complicated by using a whole lot of steps.

gee sorry i was just giving another way to do it which uses logic instead of just subbing into formulas

n u do 3 unit so u should appreciate my methods

 

[Takuya]

Originally posted by failingTheHsc
gee sorry i was just giving another way to do it which uses logic instead of just subbing into formulas

n u do 3 unit so u should appreciate my methods

Not really, because you're just subbing into formulas the same way as we're doing. You're just acknowledging that at the maximum height, y velocity is zero and using the formula

r=ut+0.5(at^2)

Note s = r = displacement. I think this syllabus actually refers to it as r, not s.

If you were to do it mathematically without Newtonian formulae then it would be a different story...