Question

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Answer 1

answer are solve down . pls see them

Explanation:

For object 1st

u1( initial velocity ) 0m/s

h1( height at which dropped) = 150 m

a1 = g m/s^2 = 10m/s^2

At 2 sec , height of object from ground

we use 2nd equation of motion . i.e

$s = \: h - ( ut + \frac{1}{2} a {t}^{2} )$

$s1 = 150 - (\frac{1}{2} \times 10 \times {2}^{2} )$

$s1 = 150 - 20$

$s1 = 130$

For object 2nd

h1( height at which dropped)=100 m

u2( initial velocity) = 0m/s

a2 = g m/s^2 = 10 m/s^2

At 2 sec height of object from ground

we use 2nd equation of motion

$s \: = h- (ut + \frac{1}{2} a {t}^{2} )$

$s2 = h2 - ( \frac{1}{2} \times 10 \times {2}^{2} )$

$s2 = 100 - 20$

$s2 = 80m$

Now difference in their heights are

130 - 80 = 50 m

we earlier see that formula for their heights are

$s1 = h1 - \frac{1}{2}a {t}^{2}$

$s2 = h2 - \frac{1}{2} a {t}^{2}$

So s1 - s2 =h1 -h2

= 50 m

so the difference in their heights doesn't depends on time

or we can say that

$s1 - s2 \: ∝ \: {t}^{0}$

hope it helps u . pls mark my answer brainleist