HELP M1 Questions!!

5) Nathtan hits a tennis ball straight up into the air from a height of 1.25m above the ground. The ball hits the ground after 2.5s. Assuming g =10 find)
i) speed nathan hits the ball = 12
ii) greatest height above ground reached by ball = 8.45m
iii) the spped the ball hits the ground = 13

iv) how high the ball bounces if it loses 0.2 of its speed on hitting the ground.
for this last Q i used u= 13-0.2 = 12.8 v=0 a=-10
then used

v^2 = u^2 + 2ax

and got

x

to be 8.192m. but book says it is 5.41m, but the book isnt always right as i have discovered in the past so does anyone know if im right? thanks in advance

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Reply 1

thedemon13666

i got the same value

is that all the information given?

Are your values for the previous parts of the question right in the book?

Reply 2

amh189

yeh all other values i calculated previously are correct and the same in the book except for last part where the book says 5.41 but i still get 8.192 thts all the info given in the Q.

there is also another Q where i think im right but not according to the book.
8) A abll is dropped from a tall building and falls with acceleration of magnitude 10. The distance between floors in the block is constant. The ball takes 0.5s to fall from the 14th to the 13th floor and 0.3s to fall from the 13th to th 12th floor. What is the distance between floors?
i got 3.2m but book says 3m

Reply 3

thedemon13666

i got x = 3

can you post your working?

Reply 4

thedemon13666

Hey ive also realised now

the book is right for the first question

when it means it loses 0.2 of its speed it means 20%

see if you can get the answer now

Reply 5

amh189

i think i have done this Q wrong tryin to work it out now.
i assumed u=0 for the first part of the motion
then used s = ut + 0.5 at^2 where u = 0 and t = 0.8
but i know its completely wrong

Reply 6

thedemon13666

ah no thats the trick to this question

you cant just assume that u = 0 just because it doesnt have a speed

but really it isnt useful to talk about u in this question

for it falling past the first window it may be useful to think of the equation:

s = vt - \frac{1}{2}at^2

and then can you see the link between the final speed of the object falling past the 14th fall and the initial speed of the object falling past the 13th floor?

Reply 7

amh189

yeh thts a good point it says 0.2 it doesnt have any units after the 0.2 so it prob does mean 20% thanks got the answer to that
now just Q8 and Q9

9) Two clay pigeons are launched vertically launched upwards from exactly the same spot at 1s intervals. Each clay pigeon has initial spped 30 and acceleration 10 downwards. How high above the ground do they collide?

i used

s= ut +0.5at^2

for 1st pigeon i got

s = 30t-5t^2

and for 2nd i got

s=30(t-1) -5(t^2-1)

then put them equal to each other, but after simplifying 2nd equation you also get 30t and 5t^2 so they cancel eah other out

Reply 8

amh189

i know the final speed v of the object past the 14th window = initial speed u of the object past the 13th window.

so i got

s=0.5v - 1.25

and

s=0.3u + 0.45

but if i put them equal to each other i have 2 unknowns in the equation.

do simulatneous equations come into this question by any chance?

Reply 9

thedemon13666

i can see the main problem:

it should be

x = 30(t-1) - 5(t-1)^2

what is the answer?

Reply 10

thedemon13666

yes simultaneous equations come in very handy for the other question :smile:

Reply 11

amh189

answer to Q9 is 43.75 m
can it not be simplified to

s = 30(t-1) - 5(t^2 + 1)

sorry should have put +1 in the first eqn cos -1^2 = +1

Reply 12

thedemon13666

no i see what your trying to do

but

(t-1)^2 = t^2 - 2t + 1

remember your polynomial work?

Reply 13

amh189

oh silly mistake when u square a brakcet u always get 3 terms thanks got the answer to that Q too and to Q 8

for Q8 i used

s = ut + 0.5at^2

s = 0.5u + 1.25

2s = 0.8u + 3.2

0.4u + 1.6 = 0.5u + 1.25

0.35 = 0.1u

u = 3.5

then substitue 3.5 into

s = ut + 0.5at^2

and u get s = 3.

thanks for all your help really appreciate it

Reply 14

thedemon13666

you know how you got u = 3.5

what speed is that supposed to be?

final speed falling past the 14th floor/ initial speed falling past the 13th floor

or is it the initial speed falling past 14th floor

BECAUSE i got the final speed falling past the 14th floor/ initial speed falling past the 13th floor as 8.5

here is my working:

for falling past the 14th window:

s = 0.5v - 1.25

then as v for falling past the 14th window = u for falling past the 13th window

s = 0.3v + 0.45

from combining the equations makes v = 0.85

and this s = 3

I can see how you did it, wouldnt of thought of that haha

did you manage the one about the clay pigeons now?

Reply 15

amh189

u = 3.5 is the initial speed falling past 14th floor
yeh i also get v = 8.5 if you use v= u + at and u =3.5 t= 0.5 and a=10
but i didnt calculate v i just used the u vlaue how did u get your answer?

i based my working on a similar Q previously in the book which is a worked example

Reply 16

thedemon13666

my workings above lol

i see now you just doubled s and added the times

did you manage the clay pigeon question?

Reply 17

amh189

i understand your method here look at this diagram it might help see where i got my working from

first i made 1 equation using t=0.5s

then i made an eqaution using 2s t=0.3+0.5 = 0.8s

does that help?

untitled.bmp349.3KB

Reply 18

amh189

yep thts exactly what i did i just drew a diagram first and saw the 2s part yeh managed clay pigeon one, really shouldnt have made such a basic mistake!! thanks for your help!