Question:
A ball is projected vertically upward with a speed of $50 \mathrm{~m} / \mathrm{s}$ Find
(a) The maximum height
(b) The time to reach the maximum height
(c) The specd at half the maximum height. Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$
Solution:
$\mathrm{u}=50 \mathrm{~m} / \mathrm{s} ; \mathrm{v}=0 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-\mathrm{g}$
(a) $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
$0^{2}=(50)^{2}+2(\mathrm{~g}) \mathrm{s}$
$s=125 m$
(b) $v=u+a t$
$0=50$-gt
$\mathrm{t}=5 \mathrm{sec}$
(c) Speed at s= $\stackrel{125}{2}=62.5 \mathrm{~m} ; \mathrm{u}=50 \mathrm{~m} / \mathrm{s}$; $\mathrm{a}=-\mathrm{g}$
$v^{2}=u^{2}+2 a s$
$=(50)^{2}-2(\mathrm{~g})(62.5)$
$\mathrm{v} \approx 35 \mathrm{~m} / \mathrm{s}$