Question

Holla! Solve the attachment NO SPAM​

Answer 1

Answer:

$\large\color{green}\underline{\underline{Given:}}$

▪︎Intial velocity of the ball or 'u' = 19.6m-¹

▪︎Time taken by the ball to reach the ground or 't'= 5 sec

▪︎Accelertation due to Gravity or 'a' = 9.8 m-²

$\large\color{blue}\underline{\underline{To\:find:}}$

(i)The height of the tower = ??

(ii)Velocity of the ball reaching the ground = ??

$\large\color{red}\underline{\underline{Taken:}}$

$t=\frac{u}{g}$

$Size\:of\:tower=ut+\frac{1}{2}g$

$velocity=u+gt$

$\large\color{orange}\underline{\underline{Concept:}}$

Equation 1.

• First find time taken to reach the ball at max height by taking the first formula .

• Then find the height of the tower by taking the second formula .

Equation 2.

• To find the velocity take the third formula .

$\large\color{purple}\underline{\underline{Solution:}}$

Equation 1 :

$Time=\frac{19.6m-1}{9.8m-2}$

$\boxed{Time=2\:sec\:ball\:takes\:to\:reach\:the\:max\:height}$

So , the ball takes 2 seconds to reach max height and two seconds back to the tower . So it takes only one second to reach the ground .

So , there is only one second to calculate the size of the tower .

$Now,find\:the\:size\:of\:tower$

$Taken , ut+\frac{1}{2}g$

$\mapsto{19.6×1+\frac{1}{2}×9.8m}$

$\mapsto{19.6+4.9}$

$\therefore\boxed{24.5\:m\:is\:the\:height\:of\:the\:tower}$

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Equation 2 :

$Now,find\:the\:velocity\:reaching\:the\:ground$

From the zero intial velocity at time is 2 sec to the ground where the time is 5 sec . Now , the time taken 3 sec from max . So , the height to the ground .

Now ,

$v=u+gt$

$v=0+9.8×3$

$v=0+29.4$

$\therefore\boxed{velocity=29.4m/s}$

HOPE IT HELPS YOU

Answer 2

$\color{purple}\large\underline{\underline{To\:Find:}}$

$\mathtt{The\:height\:of\:the\:tower.}$

$\mathtt{The\:velocity\:after\:reaching\:the\:ground.}$

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$\color{blue}\large\underline{\underline{Concept:}}$

$\text{The height of the ball can}$ $\text{be found by using the formula.}$

$\text{The velocity after reaching the }$$\text{ground will be taken as positive is it is falling under he gravity...}$

$\text{The velocity of the ball after reaching}$ $\text{the ground is it's final velocity.}$

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$\color{orange}\large\underline{\underline{Given:}}$

$\mathtt{Initial\:Velocity = 19.6m\:s^{-1}}$

$\mathtt{Time = 5s}$

$\mathtt{Accelaration = 9.8m\:s^{-2}}$

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$\color{red}\large\underline{\underline{We\: know:}}$

$\color{blue}{\mathtt{\rightarrow s = ut + \dfrac{1}{2}at^{2}}}$

Where,

$\mathtt{a = accelaration}$

$\mathtt{t = time\:taken}$

$\mathtt{s = height}$

$\color{blue}{\mathtt{\rightarrow v = u - gt}}$

Where,

$\mathtt{v = final\:velocity}$

$\mathtt{u = initial\:velocity}$

$\mathtt{g = accelaration\:due\:to\:gravity}$

$\mathtt{t = time\:taken}$

$\color{blue}{\mathtt{\rightarrow v^{2} = u^{2} - 2gh}}$

Where,

$\mathtt{v = final\:velocity}$

$\mathtt{u = initial\:velocity}$

$\mathtt{g = accelaration\:due\:to\:gravity}$

$\mathtt{h = height}$

$\color{blue}{\mathtt{\rightarrow h_{max} = \dfrac{u^{2}}{2g}}}$

Where,

$\mathtt{h = height}$

$\mathtt{u = initial\:velocity}$

$\mathtt{g = accelaration\:due\:to\:gravity}$

$\color{blue}{\mathtt{\rightarrow h = \dfrac{1}{2}gt^{2}}}$

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$\color{navy}\large\underline{\underline{Taken:}}$

$\text{let the height be h.}$

$\text{let the final velocity be v.}$

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$\color{purple}\large\underline{\underline{Solution:}}$

$\mathrm{At\:the\:highest\:point, v = 0}$

$\textit{For upward journey , from relation:}$

$v^{2} = u^{2} - 2gh$

$\Rightarrow 0^{2} = u^{2} - 2gh$

or

$h_{max} = \dfrac{u^{2}}{2g}$

$\Rightarrow h_{max} = \dfrac{19.6^{2}}{2 \times 9.8}$

$\Rightarrow h_{max} = \dfrac{19.6 \times 19.6}{19.6}$

$\Rightarrow h_{max} = 19.6 m$

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$\mathrm{If \:the\:ball\:takes\:time\:t_{1} to\:go\:to}$$\mathrm{go\:to\: highest\:point\:from\:the}$$\mathrm{top\:of\:building,\:then\:for\:upward}$$\mathrm{journey,\:from\:relation}$

$v = u - gt_{1} , [v = 0]$

$\Rightarrow 0 = 19.6 - 9.8t_{1}$

$\Rightarrow -19.6 = - 9.8t_{1}$

$\Rightarrow \dfrac{-19.6}{-9.8} = t_{1}$

$\Rightarrow 2 sec = t_{1}$

$\textsf{Time taken to reach the ground}$

$\Rightarrow Total\:time - Time\:taken\:for\:Upward\:journey$

$\Rightarrow t_{2} = 5 sec - 2 sec$

$\Rightarrow t_{2} = 3 sec$

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$\mathtt{Final\:Velocity}$

$\textit{Now for downward journey from the highest}$$\textit{point , if the ball takes time to reach the ground}$

$We know$

$\mathrm{u = 0 , g = 9.8ms^{-2} , t_{2} = 3 sec}$

$From\:relation$

$v = u + gt$

$\Rightarrow v = 0 + 9.8 \times 3$

$\Rightarrow v = 29.4ms^{-1}$

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$\mathtt{Height\:of\:the\:tower}$

$\textit{From the relation}$

$s = ut_{2} + \dfrac{1}{2}at_{2}^{2} [u = 0]$

$\Rightarrow s = 0 \times 3 + \dfrac{1}{2} \times 9.8 \times 3^{2}$

$\Rightarrow s = 4.9 \times 9$

$\Rightarrow s = 44.1 m$

$\Rightarrow Height\:of\:tower = Total Hight - height_{max}$

$\Rightarrow Height\:of\:tower = (44.1 - 19.6)m$

$\therefore Height\:of\:tower = 24.5 m$

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