Question

From the top of a vertical tower 248 m high a stone is projected vertically downwards with a speed of 8ms^-1. After x seconds, another stone is projected vertically upwards from the level of the base of the tower with a speed of 25ms^-1.Given that the two stones are at the same height moving in the same direction, 6 second after the projection of the first stone, calculate
a) The value of x
b) The velocity of the second stone at this instant

Answer 1

s1 = -8 t - (1/2) (9.8 m/s^2) t^2
s2 = 25 (t-x) -(1/2)(9.8 m/s^2)(t-x)^2

s1 = s2 at t=6 substitute and solve for x.

v2 = 25 - (9.8 m/s^2)(t-x)

Answer 2

why is it -8

Answer 3

correction: x = x0 + v0 t + (1/2) a t^2

Answer 4

it's going down

Answer 5

so the distance that the first stone covered is 228m

Answer 6

so 248-228=20

Answer 7

so 20=25(6-x)-1/2(10)(6-x)^2=-5x^2+35x-50=0

Answer 8

x= either 2 or 5 so 6-2=4 seconds

Answer 9

use v^2=u^2+2as so (25*25+2(-10)(20)^1/2

Answer 10

i checked out the answers in my mechanics book and it appears to be 2 and 14.2

Answer 11

when i replaced the t with 6 in the last part of your working i have found out the answer to be 458

Answer 12

anyways thank you for your help

Answer 13

and i know that it decelerates, what iam saying here is that when i substituted t by 6 the answer is not still the same as in the markscheme of my book

Answer 14

corrected position function
s(t) = -4.5t^2 - 8t + 248

Answer 15

@ganeshie8 @agent0smith

Answer 16

evaluate at s(6) and this should reveal to you where the stone is

Answer 17

@ganeshie8 how would you set up the second stone with the time delay?
I was just going to do a straight forward
s(t) = -4.5t^2 + 25t + 0
then evaluate at s(6) also then figure out where the position is supposed to be
obviously this will be different because it has a greater initial velocity and it is decelerating as it goes up.
after evaluating at s(6), solve for t at the distance where the two stones meet
add (or subtract) the t to 6

Answer 18

\(s(t_1) = -4.5t_1^2 - 8t_1 + 248\)

\(s(t_2) = -4.5t_2^2 + 25t_2 + 0\)

Answer 19

and since second stone is fired x time after first stone,

\(t_2 = t_1 + x\)

Answer 20

ur method also works... but in the background u r using t2 is lagging i think...

Answer 21

I didn't know how to squeeze the delay into the second function
I was thinking that the second function be evaluated how long it will take for it to reach the same position as the first stone's s(6) position
6-t = x

Answer 22

i made a typo earlier. corrected below :
\(t_2 = t_1-x\)

so that t_2 becomes 0 when t_1 = x,
implying that second stone starts x seconds after the first stone