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
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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
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