Question

a stone is dropped from the top of a tower and it strikes with 3km/hr against the ground the ground.another stone is thrown vertically downwards from the same top of tower with a velocity 4km/hr. Its velocity when it strikes the ground will be

Answer 1

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Answer 2

Answer:

The velocity when it strikes the ground is 5 km/hr.

Explanation:

Given that,

Downward velocity = 3 km/h

Vertical velocity = 4 km/h

For stone first, (free fall)

We need to calculate the height when drop the stone

Using equation of motion

$v^2-u^2=2gs$

Where, s = height

Put the value into the formula

$h =\dfrac{3^2}{2g}=\dfrac{9}{2g}$

For second stone, (through vertically)

We need to calculate the maximum height

Using equation of motion

$v^2-u^2=2gs$

Where, s = maximum height from top of tower

$h'=\dfrac{16}{2g}$

We need to calculate the total height

$H = h+h'$

$H =\dfrac{25}{2g}$

Now, velocity of second stone strike the ground

$v=\sqrt{2gH}$

Put the value into the formula

$v=\sqrt{2g\times\dfrac{25}{2g}}$

Therefore, v = 5 km/h

Hence, The velocity when it strikes the ground is 5 km/hr.