Question

A stone is dropped from a certain height after falling
for 5sec. the stone breaks through pane of glass and
instantly loses half of its velocity. If the stone
then takes one more second to reach the ground
Determine the height of glass above ground
g=10m/s²​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Time when it reaches the glass pane = 5 seconds.
• It loses ½ velocity after it strikes the glass pane.
• Initial velocity (u) = 0 .
• g = 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Velocity through it strikes the glass pane (v) = ?.

Applying first kinematics equation,

$\large{\boxed{\tt v = u + at}}$

Substituting the values,

$\large{v = 0 + 10 \times 5}$

$\large{\tt v = 10 \times 5}$

$\large{\boxed{\tt v = 50 \: m/s}}$

Therefore it strikes the glass pane with a velocity of 50 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, as mentioned in question,

The stone loses ½ velocity after striking,

Therefore, after passing through the glass pane the velocity will be,

$\large{\boxed{\tt v' = \dfrac{v}{2}}}$

Substituting,

$\large{\tt v' = \dfrac{50}{2}}$

$\large{\tt v' = \dfrac{\cancel{50}}{\cancel{2}}}$

$\large{\boxed{\tt v' = 25 \: m/s}}$

This is the final velocity after passing through glass pane.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Applying second kinematic equation,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

Note:-

Here u = v'.

$\large{\tt S = 25 \times 1 + \dfrac{1}{2} \times 10 \times (1)^2}$

As it is mentioned in question,

That it takes further 1 second to reach the ground,

Therefore, t = 1 second.

Simplifying,

$\large{\tt S = 25 + \dfrac{1}{\cancel{2}} \times \cancel{10} \times 1}$

$\large{\tt S = 25 + 5 \times 1}$

$\large{\tt S = 25 + 5}$

$\huge{\boxed{\boxed{\tt S = 30 \: meters}}}$

So, the Height of the glass above the ground is 30 meters.

$\rule{300}{1.5}$

Answer 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Time Taken,t = 5s

• Acceleration due to Gravity,g = 10 m/s²

The stone starts from rest » u = 0 m/s

Using the Relation,

$\huge{\boxed{ \sf{v = u \: + \: at}}}$

Putting the values,we get:

$\sf{v = ( 10 \times 5)} \\ \\ \huge{\rightarrow \: \rm{v = 50 \: ms {}^{ - 1} }} \:$

The velocity of the stone before colliding with the pane is 50 m/s

• Since,the stone loses half its velocity after hitting the pane

Let the velocity after impact be "b"

$\sf{b = \frac{v}{2} } \\ \\ \rightarrow \: \sf{b = \frac{50}{2} } \\ \\ \huge{ \rightarrow \: \underline{ \boxed{ \sf{b = 25 \: ms {}^{ - 1} }}}}$

Now,

• The total time taken to reach the ground becomes 1 seconds after the impact

Using the Relation,

$\huge{\sf{s = ut \: + \frac{1}{2}at {}^{2} }}$

Putting the values,we get:

$\implies \: \sf{s = (25)(1) + \frac{1}{2}(10)(1) {}^{2} } \\ \\ \implies \: \sf{s = 25 \: + \: 5} \\ \\ \huge{\implies \underline{\boxed{\sf{s = 30m}}}}$

Thus,the height from which the stone is projected is 30m