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

Explanation

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`