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Motion in One Dimension — PHYSICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9PHYSICSMotion in One Dimension3 Marks
Question
A stone thrown vertically upwards, takes 4 s to return to thrower. Calculate
(i) initial velocity of the stone
(ii) maximum height attained by stone. (Take $g =10 ms^{-2}$ )
✓

Answer

A stone thrown vertically upwards, takes 4 s to return to thrower, $\Rightarrow$ Time taken by the stone to reach maximum height $=t=\frac{4}{2}$
$
t=2 s
$
Initial velocity of stone $= u =$ ?
Maximum height attained by stone $=h=$ ?
Acceleration $=a=-g=-10 ms^{-2}$
(i) $v=u+a t$
Velocity of stone at heighest point $=v=0$
$\begin{array}{l}
0=u+(-10) 2 \\
u=10 \times 2=20 ms^{-1}=20 ms^{-1}
\end{array}$
$\begin{array}{l}(ii) \quad v^2-u^2=2 a S \\
(0)^2-(20)^2=2(-10) h \\
-20 h=-400 \\
h=\frac{400}{20}=20 m
\end{array}$

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