Question

plz solve question 6
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Answer 1

Answer:

V^2 = U^2 +2aS

   V^2 = 2*9.8*50

   V^2 =  980

 After opening paracheuit

 V^2 = U^2 - 2*2*S

 2*2*S = 980 - 9

    S  = 973 /4

    S = 243 M

TOTAL DISPLACEMENT

      = 50 + 243  = 293 METER

   

Explanation:

Answer 2

Correct Questions—

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s² . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ? (Given g=9.8 m/s² approximately)

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$\huge\sf\underline{Answer}$

$\bf\huge\boxed{293 m}$

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Solution —

The speed of the person just before the parachute opens is:

$u = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 50}$

$u = \sqrt{980}$

Now

When the parachute opens and descends,

Applying NEWTON'S LAW

$v {}^{2} - u {}^{2} = 2as$

putting the value

$9 - 980 = - 2 \times 2 \times h$

Here a is negative as perticles decelerates

$h = \frac{971}{4}$

$h = 242.75m$

Therefore, The total height of fall is

242.75 + 50 = 292.45 m

=293 m (approx)