A ball is thrown vertically upwards with a speed of 29ms^-1. It hits the ground 6 seconds later. By modelling the ball as a particle find the height above the ground from which it was thrown.

how???

many thanks,

Oli

v=u+at
So, 0=29+(-)9.81*x
-29/-9.81= t = 2.956s for the upward part of the journey.
For the downward bit:
t=6-2.956 = 3.04s to the ground.
Distance for the downward bit [i.e the max height]:
s=ut+.5at^2
=0*3.04 + .5x9.81x3.04^2
= 45.33m

Count the upward part as downwards to make it easier [make everything positive i.e downwards. as everything accelerates due to gravity at the same rate as it de-accelerates due to gravity]
s=.5(u+v)t
=.5*29*2.956
=42.86m

45.33-42.86 = 2.74m

Quite how you're supposed to model that as a particle im not sure

Quote from J@tko :

v=u+at
So, 0=29+(-)9.81*x
-29/-9.81= t = 2.956s for the upward part of the journey.
For the downward bit:
t=6-2.956 = 3.04s to the ground.
Distance for the downward bit [i.e the max height]:
s=ut+.5at^2
=0*3.04 + .5x9.81x3.04^2
= 45.33m

Count the upward part as downwards to make it easier [make everything positive i.e downwards. as everything accelerates due to gravity at the same rate as it de-accelerates due to gravity]
s=.5(u+v)t
=.5*29*2.956
=42.86m

45.33-42.86 = 2.74m

Quite how you're supposed to model that as a particle im not sure

according to the book the answer's 2.4 thanks for the help though i'll leave that one blank and get my "teacher" to explain it to me... god further maths is hard.

Quote from oli17 :

according to the book the answer's 2.4 thanks for the help though i'll leave that one blank and get my "teacher" to explain it to me... god further maths is hard.

I did round everything quite majorly

BTW is that M1? Doing that this year - will be fun

Quote from J@tko :

I did round everything quite majorly

BTW is that M1? Doing that this year - will be fun

yep. "gay" doesn't begin to describe it

p.s. it's schoolboy speak, don't be offended oldies!

Quote from oli17 :

yep. "gay" doesn't begin to describe it

p.s. it's schoolboy speak, don't be offended oldies!

Oh joy - I've got M1 AND M2 this year

At least its better than bloody statistics. God they were awful.

I was good at Maths when you actually add up numbers, when algebra was thrown in i didn't do quite as well, watching you work that out gave me a brain fart.

It's much easier to just use s = ut + 0.5at^2

s = -29x6 + 0.5x9.81x36
= 2.58

The same answer as J@kto's working should have given but with no rounding errors.

You don't even really need to use SUVAT - if acceleration due to gravity is 9.81m/s it's easy to work it out.

EDIT:

29 ÷ 9.81 = 2.956... = Z = After Z seconds it's stopped at the maximum height, velocity of 0.

Assuming an average speed of 14.5 m/s (average of 29 & 0) between original point and max height, starting point is 42.864... M below the max height.

The downward journey must therefore take 3.043... seconds (6-Z), which would mean it would hit the ground at 29.86 m/s. Again assuming an average speed of 14.93m/s we can tell the distance from the max height to the floor is 45.444... M.

Subtracting the original point -> max height distance from this we get...


2.58 M.

Simple.

(and i got a shit grade in Maths)

To get exactly 2.40m, round the solutions of each equation to 3sf.

Time taken to reach peak height, t
= (v-u)/a
= (0-29m/s)/-9.81m/s^2
= 2.96sec (to 3sf), therefore 3.04sec left

Distance covered upward, su
= (v^2 - u^2)/2a
= (-29m/s)^2 / 2*-9.81
= 42.9m (to 3sf)

Distance covered downward, sd
= ut+1/2(at^2)
= 0*3.04 + 1/2(9.81*3.04^2)
= 45.3m (to 3sf)

Height above the ground from which the ball was thrown, h
= sd - su
= 45.3m-42.9m
= 2.4m (exactly)

Edit: is this for GCSE level? I'd be uncomfortable rounding intermediate results like that in GCE A-levels and beyond even if the question told me to. My physics teacher would have gotten a right fit if anything but the final answer was rounded any more than to 3dp

Quote from J@tko :

Quite how you're supposed to model that as a particle im not sure

That just means assume it has zero volume (ie, it's a point mass) so you can ignore things like air resistance.