A stone dropped from a height of 100m takes 't' sec to reach the ground, at t/2 sec, height from the ground

Question

a)50m
b)25m
c)75m
d)10m

Answer 1

The correct answer is the third choice. (c) 75 m.

Many may go for (b) 25 m as correct as they calculate the fall from top in t/2 seconds, which is 25m. But the height above the ground is asked. It is 100-25 = 75 m.

The distractors (the choices) given here, are not the best alternatives. In an MCQ the choices should be such that one cannot use mere logic (not physics) and get the correct answer. Here, one can logically guess that the answer is >50 m. There is only one choice given with ans >50. So, (c) 75 m is correct. This is not good. At least 2 or 3 choices given should have been >50m.

The qick way of getting at the answer using good physics is as follows.

S = ut + 1/2 at*2 = 1/2 at*2 as initial velocity u = 0.

For time t/2, S1 = 1/2 g (t/2)*2 = 1/4 of 1/2gt*2 = /4 of 100 m = 25 m. Hence height above the ground is = 100 - 25 = 75 m.

Note that one need not assume the value of g nor calculate the value of time t.

Answer 2

b) 25m
using s=ut+1/2 at^2
assuming u to b 0 and g=10m/s^2
100=10.t^2/2
from the above t^2=20
implies t^2/4=5
?=1/2.10.5
ans=25m

Answer 3

assuming in time t it travels 100m. then using s=1/2at^2 is for 100m. so s1=1/2a(t/2)^2=(1/2at^2)/4=s/4=100/4=25m is the distance it has travelled or it is 75m from the ground.

Answer 4

y=1/2gt^t y=100 and g=10 then t=20 sec and t/2=10 ->>y=1/2*10*10=50

Answer 5

d=1/2 a*t^2
100=1/2*a*t^2
t^2=200/a
x=1/2 a*(t/2)^2
x=1/2*a*1/4 *t^2
x=1/2*a*1/4*200/a=25 meters
distance from the ground=100-25=75 meters

Answer 6

25 m using formula s= ut +1/2 at^2

Answer 7

75 meter

logically because it is based on physics that velocity changes due to gravitation and time , so velocity in first "t/2" seconds is less then in total "t" seconds. [ d/dx : rate of change]

Therefore, the velocity of the particle increases with time factor totally attributable to added distance covered in later added "t/2" seconds [which should be more than 50 meters]. It clearly follows that distance covered in first "t/2" is less than distance covered in later "t/2" seconds.

Therefore, the first "t/2" seconds might cover only 25 meters and there lies the answer 75 meters.

My answer is based only on physics logic and not actual calculations.

Regards,

Answer 8

75 m

Answer 9

Difficult to believe there are wrong answers above...

Assuming g = 10 and not 9.81...
and e = 100m

let's find t first:
e = g.t² / 2
or t = sqr(2.e/g) = sqr(200/10) = 4.4721

Now, for half time:
t/2 = 2.2306

find the distance after t/2:
e = g.t² / 2 = 10. 2.23 . 2.23 / 2 = 24.877 ~25

Answer 10

h1=100m h2=x
t1=1sec
t2=2sec
h1/h2 = t1/t2
h2= t1 h1/t2
x=1*100/2
x=100/2
x=50m
h2=50m
ans=50m