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A projectile is thrown with velocity u making angle $\theta$ with vertical. It just crosses the tops of two poles each of height h after 1 second respectively. The maximum height of projectile is

$\begin{array}{1 1}(A)\;9.8 \;m\\(B)\;19.6\;m \\(C)\;39.2\;m\\(D)\;4.9 \;m \end{array} $

1 Answer

$t_H =\large\frac{v_0 y}{g} $ and
$H= \large\frac{v^2_0 y}{2g} =\large\frac{(gtH)^2}{2g} $$=\frac{1}{2} gt^2H$
Here $t_H =\large\frac{(1+3)}{2} $$=2 s$
Hence $H= \large\frac{1}{2}$$(9.8) (2)^2 =19.6\;m$
Hence B is the correct answer.
answered Jul 14, 2014 by meena.p

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