Question

A ball is allowed to fall from rest from height h. If
it travels
9/25 of total height in last second of its
fall then ball will hit ground with speed
(g = 10 m/s2)​

Answer 1

Answer:

50m/s

Explanation:

1/2g$t^{2}$(9/25)=g/2(2t-1)

9$t^{2}$-50t+25=0

t=5s

v=gt=10*5=50m/s

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Answer 2

A ball is allowed to fall from rest from height h and it travels 9/25 of total height in last second of its fall.

We have to find the speed of the ball by which it hit the ground.

A/c to question, last second ball travels 9/25 of total height.

we assume the time taken by the ball to reach the ground is t.

∴ distance travelled by the ball in (t -1) sec

= $h-\frac{9}{25}h=\frac{16h}{25}$

initial velocity of the ball is zero. i.e., u = 0

using formula,

          $s=ut+\frac{1}{2}at^2$

so the ball reaches the ground in t sec is given by,

 $h=\frac{1}{2}\times10t^2$ ...(1)

also the ball reaches 16h/25  in (t - 1) sec.

here, s = $\frac{16h}{25}$ , u = 0 , a = 10 m/s² , t = (t - 1) sec

⇒ $\frac{16h}{25}$ = $\frac{1}{2}\times10(t-1)^2$ ....(2)

from equations (1) and (2) we get,

⇒ $\frac{25}{16}=\frac{t^2}{(t-1)^2}$

⇒ $\frac{4}{5}=\frac{t}{t-1}$

⇒ t = 5sec

now the velocity of the ball by which it strikes the ground is given by,

v = u + at

here, u = 0 , g = 10 m/s² and t = 5 sec

∴ v = 10 × 5 = 50 m/s

Therefore the velocity of the ball would be 50 m/s.