A ball is projected vertically upward with a speed of

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= $\stackrel{125}{2}=62.5 \mathrm{~m} ; \mathrm{u}=50 \mathrm{~m} / \mathrm{s}$; $\mathrm{a}=-\mathrm{g}$

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

$=(50)^{2}-2(\mathrm{~g})(62.5)$

$\mathrm{v} \approx 35 \mathrm{~m} / \mathrm{s}$

Leave a comment

None
Free Study Material