Question

please answer question no. 10( according to 5 marks)
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Answer 1

Explanation:

Initial difference in height $= (150 - 100)m = 50m$

We know that, by second equation of kinematics, $s = ut + \frac{1}{2} a {t}^{2}$

Considering,

$g = 10m {s}^{ - 2}$

Distance travelled by first body in

$2s = h1 = 0 + ( \frac{1}{2} )g {(2)}^{2} =2g = 2 \times 10 = 20m$

After 2s, height at which the first body will be $= h2 = (100 - 20)m = 80m$

Thus, after 2 s, difference in height

$(130 - 80)m = 50m$= initial difference in height

Thus, difference in height does not vary with time. So the answer is zero.

Answer 2

Answer:   plz mark as brainliest

for Ist object. u =0. t=2sec

the object is 150m abobe the ground

a =g= 10 (9.8)

s=ut+1/2at^2

s=0×2+1/2×10×2×2

s=20

AFTER 2SECOND THE DISTANCE BETWEEN OBJECT AND GROUND IS. 150-20=130M

for IInd object. u =0. t=2sec

a =g= 10 (9.8)

the object is 100m abobe the ground

a =g= 10 (9.8)

s=ut+1/2at^2

s=0×2+1/2×10×2×2

s=20

AFTER 2SECOND THE DISTANCE BETWEEN OBJECT AND GROUND IS. 100-20=80M

THE DISTANCE BETWEEN THE OBJECTS AFTER 2 SEC

===130-80=50

Explanation: