Question

A parachutist after bailing out, falls 50m without friction.When parachute opens , it decelerates at 2m/s^2. He reaches the ground with a speed of 3m/s.At what height did he bail out?

Answer 1

options are-
a) 293m

b) 111m

c)91m

d)182m

Answer 2

any body........plz help me

Answer 3

NSTSE?

Answer 4

oh!! exactly....

Answer 5

Lets say he jumped out at height h

Answer 6

ok

Answer 7

\[\large{v^2=u^2+2as}\]

Answer 8

2 x 10 x 50 = v^2
v=10root10 m/s
this is the downward speed when the parachute opens

Answer 9

^ use the above equation.

Answer 10

ok

Answer 11

now he has to fall (h-50) metres...use the same equation again

Answer 12

since u = 0 m/s

\[\large{v^2=2as}\]
s = 50 m

g = 9.8 m/s^2

Answer 13

ok then

Answer 14

only this time a = 10 - 2 m/s^2
because the parachute creates a drag force in the upward direction

Answer 15

then

Answer 16

but a is already given @him1618

Answer 17

\[\large{v=\sqrt{20 * 50} = \sqrt{1000} =10\sqrt{10} ms^{-1}}\]

Answer 18

but the net acceleration is acceleration duwe to g - accn due to parachute

Answer 19

or maybe u can take a as 2 m/s^2

try both..it only depends on the question..u should know the method

Answer 20

ok

Answer 21

so @him1618

Answer 22

\[\large{10\sqrt{10}-3=u}\]
\[\large{v=0}\]

Answer 23

@Harkirat

Answer 24

\[\large{0=(10\sqrt{10}-3)^2+2(2)(s)}\]

Answer 25

\[\large{0=818.6-4s}\]
\[\large{4s=818.6}\]
\[\large{s=24.65 m}\]

Answer 26

sorry it is s = 204.65 m

Answer 27

\[\large{s=204.65 m}\]

Answer 28

add 50 m now ...

50 + 204.65 = 254.65 m

Answer 29

so in options it is not given

Answer 30

hey!! mathslover can we meet pathak sir tomorrow...

Answer 31

Nahin ruk.

Answer 32

what about @mathslover done...

Answer 33

is a) the answer?

Answer 34

If you want to learn then dont go asking on websites...its not a good way to learn

Answer 35

\[\large{0=(10\sqrt{10})^2-4s}\]
\[\large{0=1000-4s}\]
\[\large{4s=1000}\]
\[\large{s=250 m}\]

s + 50 m = 250 m + 50 m = 300 m

Answer 36

@mayankdevnani confirm me that is a) the answer?

Answer 37

yaa

Answer 38

@mathslover

Answer 39

plzz

Answer 40

didn't you guys get this?

Answer 41

\(v^2=u^2+2as\)
\(=2*9.8*50\)
\(=980\)
This is the velocity of the man when the parachute opens.
using \(v^2=u^2+2as\) again.
\(v^2=u^2+2as\)
\(9=980-2*2*s\)
\(4s=980-9\)
\(4s=971\)
\(s=971/4\)
\(s=242.75\)
\(s=243\)
So total height
h=243+50
=293.
am I right @experimentX?

Answer 42

huh ... let ma have a look!!

Answer 43

oh k.

Answer 44

Hmm ... well the height should be less than 50. @ajprincess give another try.

Answer 45

she's right.