Question

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 approximately, did he bail out

Answer 1

Answer:

Initial velocity, u=

2gh

or, u=

2×9.8×50

=14

5

The velocity at ground, v=3 m/s

v

2

−u

2

=2as

or, 3

2

−980=2×(−2)×s

Thus, s=

4

971

, which is nearly 243 m.

Initially he has fallen 50 m

Thus total height at which he bailed out=243+50=293 m

Answer 2

Answer:

let the height be = H

Now for the first 50 meters his initial speed will be 0

and acceleration=g

using the equation: [math]v^2 = u^2 +(2as)[/math]

where u=0 and v= final speed after falling through 50 meters then

v = sqrt(1000)

This will act as the initial speed for his next part where he opens his parachute. Let that distance be x.

again using [math]v^2 = u^2 +(2as)[/math] where v^2= 9

u^2=1000 and a=-2

we get s(which is x) as 247.75 meters

H = 50 + 247.75 = 297.25

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