Question

A parachutist drops freely from an aeroplane for 10s and then the parachute opens out. Then he descends with a net retardation of 12m/s^2. If he strikes the ground with a velocity of 20m/s^2, then the height at which he gets out of the plane is​

Answer 1

Explanation:

Take a look at the attached image. Hope it helps!

Answer 2

Answer:

900m

Explanation:

initial velocity, u = 0

Time of free fall = 10s

acceleration, a=g=10m/s^2

v1 = u+at=0+10*10=100 -----at the end of free fall

s1=ut+1/2 * a * t^2 = 1/2 * 10 * 10^2 = 500m -----at the end of free ----------------------1

After parachute was deployed,

u=100m/s

a= -12 m/s^2

v=20m/s (In question, velocity that the person strikes the ground was mentioned as m/s^3, I assume it to be m/s)

v^2 - u^2 = 2as

400 - 10000 = -24s

s = 9600/24

s=400m------------------------2

Therefore total distance travelled = 1+2 = 500+400 = 900m