Vertical Motion under Gravity help needed - ASAP (1 Viewer)

[Gibbo69er]

Hey guys, got problems with two questions relating to Verticle Motion Under Gravity.

Just wondering if you could give them a go and relay the working out back to me.

Q1. A stone is thrown vertically into the air with a velocity of 10 m/s.
a) How high will it raise (Worked this one out, just added it in cos u might need it to get answer for b)
b) At what time after projection will it be 3m above ground and what will be its speed at those times?

Q2. An object is projected vertically upwards with a speed of 48 m/s. Two seconds later another object is projected vertically upwards from the same spot at the same speed. Find when and where the two objects meet.

Thanks in advance for the help.

Just incase you were wondering, the formula's are:

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

and a=-9.8

 

[shafqat]

I will get tired of saying this eventually, but you cannot use those formulas from physics, and I doubt they will give them to you in questions.
Consider only vertical motion: a = -g, integrate with respect to time and go from there.

 

[Gibbo69er]

edited by acmilan: There is no need for that, shafqat is merely stating a well known fact.

 

[shafqat]

Im sorry if I offended you. Its just that many students in the past have assumed those formulae.
a = -9.8
v = -9.8t + 10
x = -9.8t^2/2 + 10t

To work out max height, but v = 0, and sub the value for t into x.
To work out stuff at when it is 3 metres above the ground, put x = 3.

 

[Jumbo Cactuar]

For Q2 the time t is found when (x1 = x2) ie;
-4.9t₁² + v₀t₁ = -4.9t₂² + v₀t₂
-4.9t² + 48t = -4.9(t-2)² + 48(t-2)
0 = 4*4.9t + -4*4.9 - 96
t = 489/49
= 9.98 s

so x = -4.9(489/49)² + 48*(489/49)
= 5760/49
= 117.55 m

 

[nick1048]

I dont see a problem with using physics formulas if you get the correct answer!?! -_-'

 

[rama_v]

I dont see a problem with using physics formulas if you get the correct answer!?! -_-'

It doesn't show the examiner that you have any idea what calculus is, considering the topic itself is called "applications of calculus" - you have to understand the processes, not use some magic formula which only works in a few cases. For example, you cant do the last question in the 2003 hsc paper using those formulas...Not to mention that the sylabus strictly forbids the use of them.

 

[velox]

Its common knowledge. Use physics formulae in maths and you will get 0.

 

[Idyll]

I dont see a problem with using physics formulas if you get the correct answer!?! -_-'

Because chances are the first part of the question will ask you to derive the formulas starting with

d2y/dt2=-g
d2x/dt2=0

Writing down your physics formulas will get you no marks - so it's best to just learn the maths formulas. Besides an understanding of how the equations are derived and why they work makes it much easier to tackle harder problems.

 

[acmilan]

Unless they give you the formulae (in which case you dont have to do anything), it is almost guaranteed that some of the marks will be allocated to deriving the equations