A projectile is thrown with velocity u making angle θ with vertical. It just crosses tops of two po

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3-2.Motion in Plane

hard

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 $1\,s$ and $3\,s$, respectively. The maximum height of projectile is ............ $m$

A

$9.8$

B

$19.6$

C

$39.2$

D

$4.9$

Solution

$h =u \cos \theta t_1-\frac{1}{2} g t_1^2 \ldots(1)$

$h =u \cos \theta t_2-\frac{1}{2} g t_2^2 \ldots(2)$

Equating (1) and (2) and substituting the value of $t_1$ and $t_2$ we get

$u \cos \theta=19.6 \;ms ^{-1}$

Maximum height $=\frac{u^2 \cos ^2 \theta}{2 g}=19.6 \;m$

Standard 11

Physics

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