A parachutist bails out and falls 5m when his parachute opens.The deceleration now produced is 1m/s^2 .If he reaches at the ground with zero velocity,at what height did he bail out...?????
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Answers
Answer: 10 mtrs
Explanation:
According to the equation of motion in vertical plane and downward direction we have the equation as
v²= u² - 2as (a is negative due to deceleration)
As after the parachute opens he came down a distance of 5m,
s=5
a= 5m/s²
v is zero as he comes to rest finally.
Hence, u² = 2as = 2*1*5= 10m/s
Now for calculating the height after he bails out till the parachute opens, we have v or final velocity as 10 m/s
As he starts from rest his inital velocity will be zero. To find out the distance we will have
v²-u²=2gs acceleration will be acceleration due to gravity or 10m/s²
S= 100/2*10 = 5 m
Hence the total distance will be 5+5 that is 10mtrs.
Answer: 55 m
The question is divided into 2 cases. First one when he reaches the height of 5m after bailing out and second is the height which he covers after 5 m.
Explanation: The parachutist bails out and falls 5 m when parachute opens.
So,
v²=u²+2as
v²=0²+2*10*5
v²=100
v= 10 m/s
Thus, parachutist when falls 5m below, reaches a final velocity of 10 m/s which becomes the initial velocity for second case.
thus, v(1st case)= u(2nd case)= 10m/s
now, again,
v²=u²+2as
0²=10²+2*(-1)*s
0=100-2s
2s=100
s=50m
So the height at which he bailed out= (5 +50)m
= 55m