Question

. From the top of a tower of height 39.2 m a stone is
thrown vertically up with a velocity of 98 m sec'. How
long will it take to reach the ground​

Answer 1

20.39 sec

Explanation:

98 m/sec is vel upwards. and we know that gravity acts in downward direction.

So to reach the top it will take 10seconds (v =u-gt , where v= 0 since it will decelrate) .

Now it will undergo free fall from this height which is basically (39.2+ 490m)

So using S=1/2*g*t^2 on solving u get T as 10.39

so total time will be 10sec to go up + 10.39 to come down

=20.39 sec

Answer 2

$\underline{\underline{\huge{\bf{\red{Given}}}}}$

$\bullet\sf{height\: of \:tower, \:h = 39.2\: m}$

$\bullet\sf{stone\: is \:thrown\: vertically\: up\;from\:the\:tower}$

$\bullet\sf{ initial\: velocity\: of \:stone\:, u = 98\: ms^{-1} }$

$\underline{\underline{\huge{\bf{\red{To\:find}}}}}$

$\bullet\sf{Time\;taken\:by\:stone\:to\:reach\:the\:ground,t=?}$

$\underline{\underline{\huge{\bf{\red{Formula\:required}}}}}$

$\star\sf{First\;equation\:of\:motion}$

$\boxed{\sf{v=u+at}}$

$\star\sf{second\:equation\;of\:motion}$

$\boxed{\sf{s=ut+\frac{1}{2}at^2}}$

$\star\sf{third\:equation\:of\:motion}$

$\boxed{\sf{2as=v^2-u^2}}$

$\blue{\sf{\text{where u is initial velocity}}}\\\blue{\sf{\text{v is final velocity}}}\\\blue{\sf{\text{a is the acceleration}}}\\\blue{\sf{\text{t is the time}}}\\\blue{\sf{\text{s is the distance}}}$

$\underline{\underline{\huge{\bf{\red{Solution}}}}}$

$\underline{\rm{\bullet \:Finding \:time\:taken\:by\:stone\:to\:travel\:a\:height\:'p' \:above\:tower}}$

$\sf{initial \;velocity,u=98\:ms^{-1}}$

$\sf{final\:velocity,v=0\:ms^{-1}}$

$\sf{acceleration\:due\;to\:gravity,g=-\;9.8\:ms^{-2}}\\\sf{\text{(acceleration is taken as negative because object is going upwards)}}$  

$\sf{time\:taken,\;t=?}$

$\sf{distance\;travelled\;by\;stone\:above\:tower,p=?}$

Using first equation of motion

$\longrightarrow\sf{v=u+gt}$

$\longrightarrow\sf{98=0-9.8\times t}$

$\longrightarrow\sf{\purple{t=10\:s}}$

$\underline{\rm{\bullet \:Finding\:distance\:'p',\:travelled \;by\;stone\:above\:tower}}$

Using second equation of motion

$\longrightarrow\sf{2gp=v^2-u^2}$

$\longrightarrow\sf{2(-9.8)p=(0)^2-(98)^2}$

$\longrightarrow\sf{p=\frac{-(98)^2}{2(-9.8)}}$

$\longrightarrow\sf{\purple{p=490\;m}}$

$\underline{\rm{\bullet\:Finding \:time\:taken\;by\:stone\;to\;reach\;the\:ground\;from\:the\;peak\:height}}$

At this condition

$\sf{initial\:velocity,u_1=0\:ms^{-1}}$

$\sf{total\:distance\:travelled,( p + h )=529.2\:m}$

$\sf{acceleration\:due\:to\;gravity,g=9.8\:ms^{-2}}\\\sf{\text{(acceleration\:is\;taken\;as\:positive\:because\:stone\:is\:going\:down)}}$

$\sf{time\;taken\:by\;stone\;to\;reach\;ground\:from\:peak\:point ,t_1=?}$

Using third equation of motion

$\longrightarrow\sf{(p+h)=u_1t_1+\frac{1}{2}\times{g}\;{t_1}^2}$

$\longrightarrow\sf{529.2 = (0)t_1+\frac{1}{2}\times(9.8)\;{t_1}^2}$

$\longrightarrow\sf{\purple{t_1=\sqrt{\frac{529.2}{4.9}}=10.381\:s}}$

$\underline{\rm{\bullet\:Finding \:the \:total\:time\:taken\:by\:stone\:to\:reach\:the\:ground}}$

$\sf{total\:time\:required=t + t_1}$

$\boxed{\large{\purple{\sf{total\;time\:required=10\;s+10.381\;s=20.38\:s}}}}$