Question

1). A 'stone thrown vertically upwards,takes 3 s to attend the maximum height .calculate initial velocity and the maximum height attend by the stone.(take g=9.8m s‐²)​

Answer 1

To Find :

• The Initial velocity of the stone.

• The Maximum Height reached by the stone.

Given :

• Time = 3 s

• Acceleration due to gravity = 9.8 m/s²

We Know :

First Equation of Motion :

(Under Gravity)

$\bf{\underline{v = u \pm gt}}$

Where :-

• v = Final Velocity
• u = Initial velocity
• t = Time Taken
• g = Acceleration due to gravity

Third Equation of Motion :

(under Gravity)

$\bf{\underline{v^{2} = u^{2} \pm 2gh}}$

Where :-

• v = Final Velocity
• h = height
• u = Initial velocity
• g = Acceleration due to gravity

Concept :

For Finding the Initial velocity :

The value of g will be negative as the the rock is falling against Gravity.

Deriving the formula for initial velocity , when Final Velocity is equal to 0, i.e, v = 0.

Now , Using this information in the First Equation of Motion , we get :

$\implies \bf{v = u - gt} \\ \\ \\ \implies \bf{0 = u - gt} \\ \\ \\ \implies \bf{- u = - gt} \\ \\ \\ \implies \bf{\not{-} u = \not{-} gt} \\ \\ \\ \implies \bf{u = gt} \\ \\ \\ \therefore \purple{\bf{u = gt}}$

Hence , the formula for initial velocity is gt.

For the max. Height :

The value of g will be negative as the rock is moving against the gravity of vertically upwards.

Derivation of Height Maximum :-

We Know that, at the highest point , final velocity is 0 ,. i.e, v = 0.

Now using this information in the Third Equation of Motion , we get :

$\implies \bf{v^{2} = u^{2} - 2gh} \\ \\ \\ \implies \bf{0^{2} = u^{2} - 2gh} \\ \\ \\ \implies \bf{0 = u^{2} - 2gh} \\ \\ \\ \implies \bf{- u^{2} = - 2gh} \\ \\ \\ \implies \bf{\not{-} u^{2} = \not{-} 2gh} \\ \\ \\ \implies \bf{\dfrac{u^{2}}{2g} = h} \\ \\ \\ \implies \bf{h_{max.} = \dfrac{u^{2}}{2g}}$

Hence, the Formula for Height maximum is u²/2g.

Solution :

To Find the Initial velocity of the Stone :

By using the Second Equation of Motion and substituting the values in it , we get :

$\implies \bf{u = gt} \\ \\ \\ \implies \bf{u = 9.8 \times 3} \\ \\ \\ \implies \bf{u = 29.4} \\ \\ \\ \therefore \purple{\bf{u = 29.4 ms^{-1}}}$

Hence , the initial velocity is 29.4 m/s .

To find the Maximum Height of the stone :

Given :-

• Time = 3 s
• initial velocity = 29.4 m/s

By using the obtained Formula for Height maximum and substituting the values in it , we get :

$\implies \bf{h_{max.} = \dfrac{u^{2}}{2g}} \\ \\ \\ \implies \bf{h_{max.} = \dfrac{29.4^{2}}{2 \times 9.8}} \\ \\ \\ \implies \bf{h_{max.} = \dfrac{29.4 \times 29.4}{19.6}} \\ \\ \\ \implies \bf{h_{max.} = \dfrac{864.36}{19.6}} \\ \\ \\ \implies \bf{h_{max.} = 44.1 m} \\ \\ \\ \therefore \purple{\bf{h_{max} = 44.1 m}}$

Hence , the maximum height reached Is 44.1 m.