Question

A stone dropped from the tower take 3s to reach the ground. find the height of tower. g= 9.8ms-2

Answer 1

Time to reach from top to bottom=3seconds

Let the height of tower be X

g=9.8m/sec^2

$use \: equation \: of \: motion \\ s = ut + \frac{1}{2} g {t}^{2} \\ x = 0 + \frac{1}{2} 9.8 \times {3}^{2} \\ x = 4.9 \times 9 \\ x = 44.1m$
hope it will help you.

Answer 2

Answer:

44.1 m.

C. option is correct.

Explanation:

Given :

Initial velocity ( u ) = 0 m / sec

Acceleration ( g ) = 9.8 m/sec^2m/sec

2

Time ( t ) = 3 sec

First find the final velocity

we have first equation of motion

v = u + g t

v = 0 + 9.8 × 3 m / sec

v = 29.4 m / sec

Now we have third equation of motion

$\displaystyle v^2=u^2+2ghv </p><p>2</p><p> =u </p><p>2</p><p> +2gh</p><p></p><p>Put the value here we get</p><p></p><p>\begin{gathered}\displaystyle (29.4)^2=0^2+2\times9.8\times h\\\\\displaystyle (29.4)^2=19.6h\\\\\displaystyle h =\frac{864.36}{19.6} \ m\\\\\displaystyle h =44.1 \ m\end{gathered} </p><p>(29.4) </p><p>2</p><p> =0 </p><p>2</p><p> +2×9.8×h</p><p>(29.4) </p><p>2</p><p> =19.6h</p><p>h= </p><p>19.6</p><p>864.36</p><p> </p><p> m</p><p>h=44.1 m$

Thus the height of tower is 44.1 m.