Question

A Skydiver falls freely from the height of 3000 m with an acceleration of 10 ${\text{ms}}^{-2}$ and proceeds on opening the parachute after 10 s. Find the total duration of her jump, if she hits the ground at a velocity of 2 m/s?

• A. 19 seconds
• B. 59 seconds
• C. 49 seconds
• D. 25 seconds

Answer 1

The correct option is B 59 seconds

for freefall motion – First 10 seconds.
${\text{u}}_1$= 0 ${\text{ms}}^{-1}$
${\text{a}}_1$ = 10 ${\text{ms}}^{-2}$
${\text{t}}_1$= 10 s
To find unknown value (Final velocity)
using 1st eqn of motion:
${\text{v}}_1$ = ${\text{u}}_1$ + ${\text{a}}_1{\text{t}}_1$
${\text{v}}_1$ = 0 + 10 $\times$ 10
${\text{v}}_1$ = 100 ${\text{ms}}^{-1}$

To find the distance covered using 2nd eqn of motion)
${\text{s}}_1$ = ${\text{u}}_1{\text{t}}_1$ + $\frac{1}{2}$${\text{a}}_1{\text{t}}_1^2$
${\text{s}}_1$ = 0 $\times$ 10 + $\frac{1}{2}$ 10 $\times$ 10 $\times$ 10
${\text{s}}_1$ = $\frac{1000}{2}$
${\text{s}}_1$ = 500 m

After parachute opens:
${\text{v}}_1$ = ${\text{u}}_2$ = 100 ${\text{ms}}^{-1}$
${\text{v}}_2$ = 2 ${\text{ms}}^{-1}$
${\text{s}}_2$ = 3000 – 500 = 2500 m

Finding the value of new acceleration:
${\text{v}}_2^2$ – ${\text{u}}_2^2$ = 2 ${\text{a}}_2$ ${\text{s}}_2$
4 – 10000 = 2 $\times$ ${\text{a}}_2$ $\times$ 2500
-9996 = 5000 ${\text{a}}_2$
${\text{a}}_2$ = -1.9992 ${\text{ms}}^{-2}$
${\text{a}}_2$ = -2 ${\text{ms}}^{-2}$ (approx.)

Finding the time taken for second
part of journey:

${\text{v}}_2$ = ${\text{u}}_2$ + ${\text{a}}_2{\text{t}}_2$
2 = 100 + (-2) ${\text{t}}_2$
-98 = -2${\text{t}}_2$
${\text{t}}_2$ = 49 $\text{s}$

Total time taken for the entire motion:
T = ${\text{t}}_1$ + ${\text{t}}_2$
T = 59 s