Question

a stone is dropped from a height of 125 m. if the stone is acted upon by an acceleration of 10 m\sec^2 calculate when and with what velocity it will steike the ground

Answer 1

Given :

➳ Height of building = 125m

➳ Acc. due to gravity = 10m/s²

To Find :

➝ Time taken by stone to reach at ground.

➝ Final velocity at which stone touched the ground$.$

SoluTion :

Here, acceleration due to gravity continuously acts in downward direction. We can apply equation of kinematics to solve this question.

For a body, falling freely under the action of gravity, g is taken positive.

◈ Time of journey :

$\dashrightarrow\rm\:H=ut+\dfrac{1}{2}gt^2$

$\dashrightarrow\rm\:125=0+\dfrac{1}{2}\times 10\times t^2$

$\dashrightarrow\rm\:t^2=25$

$\dashrightarrow\underline{\boxed{\bf{\red{t=5s}}}}\:\gray{\bigstar}$

◈ Final velocity :

$\dashrightarrow\rm\:v^2-u^2=2gH$

$\dashrightarrow\rm\:v=\sqrt{2gH}$

$\dashrightarrow\rm\:v=\sqrt{2\times 10\times 125}$

$\dashrightarrow\rm\:v=\sqrt{2500}$

$\dashrightarrow\underline{\boxed{\bf{\blue{v=50\:ms^{-1}}}}}\:\orange{\bigstar}$

Answer 2

Given ,

Initial velocity (u) = 0 m/s

Displacement (s) = 125 m

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

We know that , the Newton's second equation of motion is given by

$\boxed{ \sf{s = ut + \frac{1}{2}a {(t)}^{2} }}$

Thus ,

125 = 0(t) + 1/2 × 10 × (t)²

125 = 5 × (t)²

(t)² = 25

t = √25

t = 5 sec

∴The stone will take 5 sec to strike the ground

Now , the Newton's third equation of motion is given by

$\boxed{ \sf{ {(v)}^{2} - {(u)}^{2} = 2as}}$

Thus ,

(v)² - (0)² = 2 × 10 × 125

(v)² = 2500

v = √2500

v = 50 m/s

∴The velocity with which the stone will strike the ground is 50 m/s