Question

abody is thrown vertically up from a wall of height 2.2m with a speed of 10m/s the distance travelled in the last second of its motion is ?

Answer 1

first :) the height doesn't even matter I think. hold on i check

Answer 2

I think it is .5102040816

Answer 3

the ans=7m

Answer 4

crap i read it wrong lol

Answer 5

do u knw the formula for distance travelled in nth second?

Answer 6

yes

Answer 7

dn=u+a(n-1/2)

Answer 8

then use that!!

Answer 9

first we have to find the total tme

Answer 10

wait a sec that n in equation is nth second ryt?

Answer 11

s

Answer 12

k then find the total time

Answer 13

ok

Answer 14

take n=total time

Answer 15

then

Answer 16

plug that into the eqn

Answer 17

leme see if i get answer as 7

Answer 18

use the formula
\[s=ut + 0.5at^2\]
s= -2.2, u =10, a=-10(assumed)
\[-2.2 = 10t -5t^2\]\[5t^2-10t-2.2=0\] solve the equation and you will get time for the whole journey t=2.2s.
distance travelled for the last second i think should mean that 1.2 - 2.2 s
To find the highest point it reaches, use
\[v^2=u^2 + 2as\] where v is the velocity when the stone reaches the highest point which is actually = 0, u=10, a=-10. sub all the values and you will get s=5m.
Now to find where does the body reach after 1.2s
\[s=ut + 0.5at^2\]
u=10,t=1.2,a=-10
sub all the values into the formula you'll get s=4.8m
So the distance travelled during the last second= (5-4.8) + 5 + 2.2 =7.4m answer may vary as i used a=-10

Answer 19

the ans=7m

Answer 20

u have to take (g=10m/s) it is there in the question

Answer 21

(5-4.8) + 5 + 2.2=7.8

Answer 22

I think that you mean the displacement travelled not the distance travelled. then it should be 4.8 + 2.2 = 7

Answer 23

no its the distance travelled in last second

Answer 24

distance travelled=7.4 displacement=7 tats the answer i found.

Answer 25

ok thanzzz