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# Solve the following :

Question:

A stone is thrown vertically upward with a speed of $28 \mathrm{~m} / \mathrm{s}$.

(a) Find the maximum height reached by the stone.

(b) Find its velocity one second before it reaches the maximum height.

(c) Does the answer pf part (b) change if the initial speed is more than $28 \mathrm{~m} / \mathrm{s}$ such as $40 \mathrm{~m} / \mathrm{s}$ or 80 $\mathrm{m} / \mathrm{s}$ ?

Solution:

(a) $u=28 \mathrm{~m} / \mathrm{s} ; \mathrm{v}=0 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-\mathrm{g}$

$v^{2}=u^{2}+2 a s$

$0^{2}=(2 s)^{2}+2(-g)(s)$

$\mathrm{s}=40 \mathrm{~m}$

(b) Velocity at one second before $\mathrm{H}_{\max }=$ Velocity after one second of $\mathrm{H}_{\max }$ So, after $\mathrm{H}_{\text {maxis }}$ attained

$\mathrm{u}=0 ; \mathrm{a}=\mathrm{g} ; \mathrm{t}=1$

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\mathrm{v}=9.8 \mathrm{~m} / \mathrm{s}$

(c) No, answer will not change. As after one second of attaining of $\mathrm{H}_{\max }$ $\mathrm{v}=9.8 \mathrm{~m} / \mathrm{s}$ only