Question

A stone is projected vertically upward with a velocity of 10ms^-1 near the edge of the cliff.Given height of cliff=20mTaking g as 10ms^-2,find
a)time taken for the stone to reach the ground
b)velocity of stone when it hits the ground
c)sketch the s-t,v-t and a-t graphs

Answer 1

a)3.2s
b)-22ms^-1

Answer 2

c)sketch the s-t,v-t and a-t graphs

Answer 3

3.2 is correct for part a

Answer 4

Yes.. @abhisar taught me 2 days back
I'm stuck with c)

Answer 5

-22 ms^-1 is also correct for part b

Answer 6

so your equation for height of stone above ground is :

\[\large s(t) = 20 + 10t + \dfrac{1}{2}(-10)t^2\]

right ?

Answer 7

yes

Answer 8

which is same as :

\[\large s(t) = -5t^2 + 10t + 20\]

Answer 9

doesn't that look like a familiar parabola facing downwards ?

Answer 10

ys

Answer 11

so basically your \(s-t\) graph is just a parabola,
can you sketch it ?

Answer 12

will try..

Answer 13

find its vertex first,
so start by changing the equation into vertex form :y = a(x-h)^2+k

Answer 14

I need to use completing to square to change it to that form

Answer 15

the*

Answer 16

yep !

Answer 17

\(\large s(t) = -5t^2 + 10t + 20\)

\(\large ~~~~~~ = -5(t^2 - 2t) + 20\)

\(\large ~~~~~~ = -5(t^2 - 2(1)t + \color{Red}{1^2}) + 20 - (\color{red}{-5*1^2})\)

\(\large ~~~~~~ = -5(t-1)^2 + 20 +\color{red}{5}\)

\(\large ~~~~~~ = -5(t-1)^2 + 25\)

Answer 18

vertex = (1, 25)
find the x/t intercepts and sketch it

Answer 19

if you want, you may also find y intercept as well to be a bit more precice

Answer 20

ok..

Answer 21

Will it be parabola facing upward or downwards ?

Answer 22

downward

Answer 23

this's the answer frm the book