Question

if a body is releated from certain height above the ground in the last second of it covers half of the total distance travelled by it then total time of its fall is​

Answer 1

Answer:

ANSWER

2

h

=

2

1

g(t−1)

2

...(1)

h=

2

1

gt

2

...(2)

t

2

+1−2t=

t

t

2

⟹t

2

−4t+2=0

g(t−1)

2

=

2

gt

2

t=

2

4±

16−8

=

2

4±2

2

t=2±

2

Now t−1>0 if t=2−

2

t−1=2

2

−1=1−

2

<0

(ii) So t=2+

2

(i) h=

2

1

gt

2

=17m

Explanation:

I hope it helps u

Answer 2

Apply 2nd equation for total time of fall ,

$:\implies \sf s=(0)t+\dfrac{1}{2}gt^2$

$:\implies \sf s=\dfrac{gt^2}{2}...(1)$

Apply 2nd equation for last second of its fall ,

$:\implies \sf \dfrac{s}{2}=(0)t+\dfrac{1}{2}g(t-1)^2$

$:\implies \sf \dfrac{s}{2}=\dfrac{g(t-1)^2}{2}$

$:\implies \sf s=g(t-1)^2$

Solve (1) ÷ (2) ,

$:\implies \sf \dfrac{s}{s}=\dfrac{\frac{gt^2}{2}}{g(t-1)^2}$

$:\implies \sf 2=\dfrac{t^2}{(t-1)^2}$

$:\implies \sf \sqrt{2}(t-1)=t$

$:\implies \sf \sqrt{2}t-\sqrt{2}=t$

$:\implies \sf t(\sqrt{2}-1)=\sqrt{2}$

$:\implies \sf t=\dfrac{\sqrt{2}}{\sqrt{2}-1}$

$:\implies \sf t=\dfrac{\sqrt{2}}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}$

$:\implies \sf t=2+\sqrt{2}$

So , Total time of fall = 2 + √2 s