Question

A particle is dropped under gravity from rest from a height h (g=9.8 m/s2) and it travels a distance 9h/25 in the last second, the height is?

Answer 1

Answer:

The height is 122.5 m.

b is correct.

Explanation:

Given that,

Distance in last second$D=\dfrac{9h}{25}$

Let us consider that a particle takes time t to reach the ground.

Using equation of motion

The total distance will be

$s= ut+\dfrac{1}{2}gt^2$

$h = 0+\dfrac{1}{2}\times9.8\times t^2$....(I)

The particle covers the distance D' in (t-1)second

$D'=h-D$

$D'=h-\dfrac{9h}{25}$

$D'=\dfrac{16h}{25}$

Using equation of motion again

$\dfrac{16h}{25}=0+\dfrac{1}{2}g(t-1)^2$....(II)

Divided equation (II) by equation(I)

$\dfrac{16}{25}=\dfrac{(t-1)^2}{t^2}$

$\dfrac{16}{25}\times t^2=(t-1)^2$

$\dfrac{4}{5}\times t=t-1$

$t = 5\ sec$

Now, put the value of t in equation (I)

The height is

$h = \dfrac{1}{2}\times9.8\times5\times5$

$h = 122.5\ m$

Hence, The height is 122.5 m.

Answer 2

We have to find numerical value of h.  Suppose, t is time of fall. We assume initial velocity for fall as zero. Then,

=>h = (1/2) g t^2……………(1)

=>h' = (1/2) g (t - 1 )^2……….(2)

Therefore , h - h' = (9h/25) = (1/2)g [ t^2 - (t -1 )^2 ].

Therefore , (9h/25) = gt - g/2.

Putting the value of h in terms of t from equation (1),

=>(9/25)(1/2)gt^2 = gt - g/2 OR

=> (9/50)t^2 -t + (1/2) = 0. OR

=>9 t^2 - 50t + 25 = 0.

Solving this quadratic equation,

t= (5/9) OR t= 5 s.

t =5 second is possible in the present, because taking t= (5/9)s, we get h= 25/9 m for g= 10 m/ s^2.

For, t=5 s,  h= (1/2)(10)(5)^2 = 125 m.