Question

A parachutist drops freely from helicopter for 20s before the parachute opens out. Then he descends with retardation of 40m/s^2. The height of the helicopter from the ground level is 2500m. When he drops out, his velocity on reaching the ground will he-man
1. 10m/s^-1
2. 10√10m/s^-1
3. 20m/s^-1
4. Zero

Answer 1

answer : option (4)

explanation : initial velocity of parachutist , u = 0

velocity of parachutist after 20s , v = u - gt

= 0 - 10 × 20 = -200 m/s

now, distance covered by parachutist during 20 sec , s = ut - 1/2gt²

= 0 - 1/2 × 10 × (20)² = - 5 × 400 = -2000m

now, remaining distance of parachutist from the ground , h = 2500 - 2000 = 500m

now using formula , v² = u² + 2gh

here, u = 200m/s, g = -40m/s², h = 500m

or, v² = (200)² - 2 × 40 × 500

= 40000 - 40000

= 0

hence, velocity of parachutist on the reaching the ground will be zero.

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

→Initial velocity

u = 0

→Now we have to find velocity of parachutist after 20s.

→Using 1st Equation of Gravitation :-

${\boxed{v = u-gt}}$

= 0 - 10 × 20

= -200 m/s

★Now we have to find the distance travelled during 20 sec :-

${\boxed{Using \:2nd \:equation \:of\: Gravitation}}$

$\tt{\rightarrow s = ut-\dfrac{1}{2}\times gt^{2}}$

$\tt{\rightarrow s = 0-\dfrac{1}{2}\times 10\times (20)^{2}}$

= - 5 × 400

= -2000m

→Now we have to find distance from the ground ,

h = 2500 - 2000

h = 500m

★Equation of gravitation we have,

${\boxed{v^{2} = u^{2} + 2gh}}$

u = 200m/s, g = -40m/s², h = 500m

→Substitute the values we get :-

v² = (200)² - 2 × 40 × 500

= 40000 - 40000

= 0

→Parachutist velocity is 0 on the reaching the ground.

Hence option (4.Zero) is correct.