Question
An object falling freely from above is in the last second its total height crosses the division of 16 / 25 then find :
(i) Total height, (ii) Time of fall.
(i) Total height, (ii) Time of fall.
✓
Answer
Suppose total height $= h$ meters and time taken to cover $= t sec$.
Distance in last second $=\frac{16}{25} h$
$(t-1) sec$. distance from
$=h-\frac{16}{25} h=\frac{9}{25} h $
From $h = ut +\frac{1}{2} gt ^2$
$u =0 \quad \because$ The object is falling freely.
$\begin{array}{l}
h=0+\frac{1}{2} \times 10 \times t^2 \\
h=5 t^2.....(1)
\end{array}$
Similarly
$\frac{9}{25} h=5(t-1)^2$..........(2)
Dividing equation (1) by equation (2) gives
$\frac{9}{25}=\frac{(t-1)^2}{t^2}$
or $ \frac{3}{5}=\frac{ t -1}{ t }$
$\begin{array}{ll}
\Rightarrow & 3 t=5 t-5 \\
\Rightarrow & 2 t=5 \Rightarrow t=2.5 sec .
\end{array}$
Putting the value of $t$ in equation (1)
$\begin{aligned}
h & =5 \times(2.5)^2 \\
& =5 \times 6.25=31.25 m
\end{aligned}$
Distance in last second $=\frac{16}{25} h$
$(t-1) sec$. distance from
$=h-\frac{16}{25} h=\frac{9}{25} h $
From $h = ut +\frac{1}{2} gt ^2$
$u =0 \quad \because$ The object is falling freely.
$\begin{array}{l}
h=0+\frac{1}{2} \times 10 \times t^2 \\
h=5 t^2.....(1)
\end{array}$
Similarly
$\frac{9}{25} h=5(t-1)^2$..........(2)
Dividing equation (1) by equation (2) gives
$\frac{9}{25}=\frac{(t-1)^2}{t^2}$
or $ \frac{3}{5}=\frac{ t -1}{ t }$
$\begin{array}{ll}
\Rightarrow & 3 t=5 t-5 \\
\Rightarrow & 2 t=5 \Rightarrow t=2.5 sec .
\end{array}$
Putting the value of $t$ in equation (1)
$\begin{aligned}
h & =5 \times(2.5)^2 \\
& =5 \times 6.25=31.25 m
\end{aligned}$
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