A body is projected from the ground at an angle of 45 ° with the...

After $2s$ the velocity is $20m{s}^{- 1}$

$⇒ {v}_{x}^{2} + {v}_{y}^{2} = {v}^{2} = 400$

$⇒ {u}^{2}\cos^{2}45 ° + {\left(usin45 ° - gt\right)}^{2} = \dfrac{{u}^{2}}{2} + {\left(\dfrac{u}{\sqrt{2}} - 20\right)}^{2} = 400$

$⇒ \dfrac{{u}^{2}}{2} + \dfrac{{u}^{2}}{2} + 400 - \dfrac{40}{\sqrt{2}}u = 400$

$⇒ {u}^{2} = \dfrac{40}{\sqrt{2}}u ⇒ u = \dfrac{40}{\sqrt{2}}m{s}^{- 1}$

Now the maximum height of the projectile is given by,

$H = \dfrac{{u}^{2}\sin^{2}\theta}{2g} ⇒ H = \dfrac{1600}{20 \times 2 \times 2} = 20m$