How far will it fall during next second? (a) H (b) 2H (c) 3H (d) 4H
4 Answers

3H. how?
initial velocity= 0 m/s. velocity after 1 sec, v = u+gt = 0+9.81*1 = 9.81 m/s
distance travelled in 1st sec = S1 = ut+0.5*g*t^2 = 0 + 0.5*9.81*1 = 4.905 m = H (as per the question)
distance travelled in 2nd second = S2 = velocity after 1sec * t+ 0.5*g*t^2 = 9.81*1 + 0.5*9.81*1 = 3 * 4.905 = 3H
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An object is released from rest and falls a distance H during the first second of time.?
The acceleration is constant.
At the beginning of the 1st second, the velocity = 0
At the beginning of the 2nd second, the velocity = vf
vf = vi + a * t
vi = 0
vf = a * t
t = 1
vf = a
Distance = average velocity * time
Average velocity = (vi + vf) ÷ 2
vi = 0
t = 1 sec
distance = vf/2
H = vf/2
H = a/2
vf = a = velocity at beginning of 2nd second
During the 2nd second, the object accelerates from an initial velocity of a to a final velocity of 2a
Distance = average velocity * time
Average velocity for 2nd second = (a + 2a) ÷ 2 = 1.5 a
Time = 1 second
Distance = 1.5 a
H = a/2
a = 2H
distance = 1.5 * 2H
distance = 3 H
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The solutions of Brite and oldprof are superb. Paul forgot about the note "for the duration of" the subsequent 2d. Paul's calculations are superb for a finished time of two seconds. In 2 seconds the article will fall a distance of 4h. Subtracting the area it has fallen contained in the first 2d, you get the area it has fallen in the course of the subsequent 2d, so d = 4h  h = 3h <      answer C

Basic physics. H= 0.5 gt^2. g= 9.81 t =1 so H = 9.81*0.5 m.. For 2 seconds t=2 so H= 0.5 X 9.81 X 4. For 3 sec, t=3 H= 0.5 9.81 X 9 and so on. It will fall for 78.48 m for 4 seconds
Source(s): High school elementary physics