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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=3s repectively. Calculate maximum height atained by it? `(g=10m//s^(2))`

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The correct Answer is:
20 m

`u sin thetaxx1-1/2 g(1)^(2)=u sin thetaxx3-1/2xxgxx(3)^(2)`
`2u sin theta=40 rArr u sin theta=20 m//s`
Max. height `=(u^(2)sin^(2) theta)/(2g)=(20xx20)/2-=20 m`
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