Question

23.A stone dropped from a certain height can reach the ground in 5 s.
It is stopped after 3 seconds of its fall and then allowed to fall
again. Find the time taken by the stone to reach the ground for the
remaining distance.
A. 2 s
B, 4 s
C. 6 s
D. 8s
E. 10 s​

Answer 1

Given:

Total time of fall from a certain height of the stone = 5 s

Time after which stone is stopped during its fall = 3 s

To Find:

Time taken by the stone to reach the ground for the remaining distance (t')

Answer:

When stone it dropped from a certain height initially:

Initial velocity (u) = 0 m/s

Time taken (t) = 5 s

Acceleration due to gravity (g) = 10 m/s²

From second equation of motion we get total height (h):

$\bf \leadsto h = ut + \dfrac{1}{2} g {t}^{2} \\ \\ \rm \leadsto h = 0 \times t + \dfrac{1}{2} \times 10 \times {5}^{2} \\ \\ \rm \leadsto h = 5 \times 25 \\ \\ \rm \leadsto h = 125 \: m$

Height covered (h') by stone in 3 s:

$\bf \leadsto h' = ut + \dfrac{1}{2} g {t}^{2} \\ \\ \rm \leadsto h' = 0 \times t + \dfrac{1}{2} \times 10 \times {3}^{2} \\ \\ \rm \leadsto h' = 5 \times 9 \\ \\ \rm \leadsto h' = 45 \: m$

So, Remaining distance/height (H) = Total height (h) - Height covered in 3 s (h')

= 125 - 45

= 80 m

Since stone is stopped after 3 s. So, it's velocity will become 0.

Thus,

Motion after stopping stone at 3 s:

Initial velocity (u') = 0 m/s

Remaining distance (H) = 80 m

Acceleration due to gravity (g) = 10 m/s²

By using second equation of motion we get:

$\bf \leadsto H' = u't' + \dfrac{1}{2} g {t'}^{2} \\ \\ \rm \leadsto 80 = 0 \times t + \dfrac{1}{2} \times 10 \times {t'}^{2} \\ \\ \rm \leadsto {t'}^{2} = \frac{80}{5} \\ \\ \rm \leadsto {t'}^{2} = 16 \\ \\ \rm \leadsto t' = 4 \: s$

$\therefore$ Time taken by the stone to reach the ground for the remaining distance (t') = 4 s

Correct Option: $\boxed{\mathfrak{B. \ 4 \ s}}$

Answer 2

• A stone dropped from a certain height can reach the ground in 5 s

Initial velocity (u) = 0 m/s

∵ dropped from height

Total height , H = ? m

Time (t) = 5 s

Acceleration due to gravity (a) = 10 m/s²

Apply 2nd equation of motion ,

$\pink{\bigstar}\ \; \sf s=ut+\dfrac{1}{2}at^2$

$:\implies \sf H=(0)(5)+\dfrac{1}{2}(10)(5)^2$

$:\implies \sf H=5(5)^2$

$:\implies \sf H=125\ m\ \;$

• This stone is stopped after 3 seconds of its fall

$:\implies \sf H=\dfrac{1}{2}(10)(3)^2$

$:\implies \sf H=45\ m$

• And this stone now allowed to fall again from rest

Remaining height (h) = 125 - 45 = 80 m

$:\implies \sf 80=\dfrac{1}{2}(10)t^2$

$:\implies \sf t^2=16$

$:\implies \sf{\bf{t=4\ s}}$

Option B