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A parachutist after bailing out falls 52.0 m without friction. When the parachute opens, she decelerates at 2.10 m/s 2 and reaches the ground with a speed of 2.90 m/s. (a) How long is the parachutist in the air? (b) At what height did the fall begin?

A  parachutist   after  bailing   out  falls  52.0  m  without  friction. When  the  parachute   opens,  she  decelerates   at  2.10  m/s2   and reaches  the ground  with  a speed  of 2.90 m/s.  (a)  How  long is the  parachutist  in  the  air?  (b) At  what  height   did  the  fall begin?

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
6 years ago
(a) Given that the parachute travels the distance say y1 pf 52.0 m without friction, the time t1 taken to travel that distance under action of gravity is given by equation (1) as:
235-1106_71.PNG
Substitute the given values,
235-1000_75.PNG
Therefore the parachutist took 3.25 s to fall by 52.0 .
235-2297_76.PNG
Substitute the given values,
235-279_77.PNG
The negative sign in acceleration accounts for the fact that the parachutist decelerates.
Therefore the time t for which the parachutist is in the air is given as:
235-1945_78.PNG
Round off to three significant figures,
t = 17.1 s
Therefore the parachutist was in the air for 17.1 s.
(b) The distance y2 travelled by the parachutist under the constant deceleration a is given using equation (3) and is derived as:
From the equation of kinematics, we have

235-569_79.PNG
Substitute the value in first equation
235-2026_80.PNG
The speed (v) is given as the product of time t1 and the free fall acceleration, therefore by substituting the same, equation (3) is derived as:
235-2242_81.PNG
The height y from which the fall began is given as:
y = y1 + y2
Substitute the given values of y1 to be 52.0 m and the calculated value of y2 in the equation above,
y = 52.0 m + 240.01 m
= 292.01 m
Round off to three significant figures,
y = 292 m
Therefore the parachutist fell from 292 m above the ground.


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