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A projectile is thrown with speed u maki...

A projectile is thrown with speed u making angle `theta` with horizontal at `t=0`. It just crosses the two points at equal height at time `t=1` s and `t=3` sec respectively. Calculate maximum height attained by it. `(g=10 m//s^(2))`

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Displacement in y direction `y=u_(y)xx1-(1)/(2)gxx(1)^(2)=u_(y)xx3-(1)/(2)g(3)^(2)rArru_(y)=2g=20m//s`
Maximum height attained `h_("max")=(u_(y)^(2))/(2g)=20m`
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