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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 3 s to attain maximum height. Calcualte
1. initial velocity of the stone
2. maximum height attained by the stone. (Take $g =9.8 ms^{-2}$ )
✓

Answer

Initial velocity of stone $=u=$ ?
At highest point, final velocity $=v=0$
$
\text { Time }=t=3 s
$
Acceleration $=a=-g=-9.8 ms^{-2}$
Maximum height attained by stone $= S = h =$ ?
$\begin{array}{l}(i) \quad v=u+a t \\
0=u+(-9.8) 3 \\
u=9.8 \times 3=29.4 ms^{-1}
\end{array}$
$\begin{array}{l}
(ii) \quad \text { Now, } S=u t+\frac{1}{2} a t^2 \\
\begin{aligned}
h & =(29.4) 3+\frac{1}{2}(-9.8)(3)^2=88.2-4.9 \times 9 \\
& =88.2-44.1=44.1 m
\end{aligned}\end{array}$

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