Question

An object is released from a height h above the ground reaches the ground in 8sec. The time taken by the object to cover the first half distance is
A)2√2s
B)4/√52s
C)4√2s
D)8√2s

Answer 1

Explanation:

h=1/2gt^2

so h'=h/2

i.e,

t= 8 seconds

h'=(1/2gt^2)/2

h'=1/4gt^2

h'=1/4×10×8×8

h'=160meters

so

h'=1/2gt'^2

160=1/2(10) t'^2

16=1/2(t'^2)

32=t'^2

so t'=√32

so t'=4√2

so t'=4√2 seconds

Answer 2

This is the concept of dropping a ball from some height.
it can follow all 3 newton equations
v = u + at -----------------( 1 )
s = ut + 1/2 at² ----------------( 2 )
v² = u² + 2as ------------------ ( 3 )

An object is released from a height h above the ground reaches the ground in 8sec
time to reach ground = 8 sec.
initial velocity = 0 m/s ( because ball is dropped ).

Time taken to cover the half distance


height covered by ball in 8 sec
s = ut + 1/2 at²
s = 0 * 8 sec + 1/2 * 10 * 64
s = 320 m
time taken to cover half the distance i.e. 160m
s = ut + 1/2 at²
160m = 0*8 + 1/2 *10 * t²
t = √32sec
t = 4√2 sec

4√2

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