Question

A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3 u the height of tower

Answer 1

Answer:

given ,

initial velocity = u

Lets take the maximum height reached by the stone as H and height of tower as h.

H = u² / 2g

lets leave it as it is .

Now , the stone falls freely .

This means ,

u = 0

v = 3u                           [ given ]

a = g

Using the third equation of motion ie S = v² - u² / 2g

since u = 0

and S = H+h

H+h = (3u)² / 2g

       = 9u² / 2g

since H = u² / 2g ,

h + u² / 2g = 9u² / 2g

h = 9u² / 2g - u² / 2g

  = 8u² / 2g

Hence the height of the tower will be 8u² / 2g .

hope it helped and please mark as brainliest:)

Answer 2

We're given,

Initial speed, u = u.

Final speed, v = 3u.

We have to find the height of the tower, say H.

Let the angle made by the stone with the horizontal be θ.

We have,

$\vec{u}=u_x\hat i+u_y\hat j\quad;\quad u_x=u\cos\theta\quad;\quad u_y=u\sin\theta\\\\\therefore\ \vec{u}=u\cos\theta\hat i+u\sin\theta\hat j$

And,

$\vec{v}=v_x\hat i+v_y\hat j\\\\v_x=u\cos\theta\quad [a=0]\\\\-v_y=u\sin\theta-gt\quad\implies\quad v_y=gt-u\sin\theta\quad[a=-g]\\\\\therefore\ \vec{v}=u\cos\theta\hat i+(gt-u\sin\theta)\hat j=\vec{3u}$

Here $-v_y$ is taken because the velocity is downwards when it reaches the ground, and $u_y$ is taken upward.

So, we know,

$\begin{aligned}|\vec{v}|=\ \ &\sqrt{u^2+g^2t^2-2ugt\sin\theta}=3u\\\\\implies\ \ &u^2+g^2t^2-2ugt\sin\theta=9u^2\\\\\implies\ \ &g\cdot\dfrac{v_y+u\sin\theta}{g}\left(g\cdot\dfrac{v_y+u\sin\theta}{g}-2u\sin\theta\right)=8u^2\ [\because\ v_y=gt-u\sin\theta]\\\\\implies\ \ &(v_y)^2-u^2\sin^2\theta=8u^2\end{aligned}$

To find H, we have only to consider vertical motion. That's why we found this.

To find H, we use  $v^2=u^2+2as\quad\implies\quad s=\dfrac{v^2-u^2}{2a}$

Here,

$s=-H$

Because the stone goes up and then down, neglecting the displacement travelled upwards.

$u=u\sin\theta\\\\v=v_y\\\\a=-g$

Now,

$\begin{aligned}&-H=\dfrac{(v_y)^2-u^2\sin^2\theta}{-2g}\\\\\implies\ \ &-H=\dfrac{8u^2}{-2g}\\\\\implies\ \ &\large\boxed{H=\dfrac{4u^2}{g}}\end{aligned}$