Aball is thrown from the ground at an angle of 45degree from the horiz

1.	Aball is thrown from the ground at an angle of 45degree from the horizontal rising to maximum height of 50m find initial velocity
Answer» GIVEN : Angle of PROJECTION = 45° Maximum height = 50m To find : The initial velocity of the ball Solution : Maximum height attained by a projectile is given by, $\boxed{ \bf H = \dfrac{u^2 sin^2 \theta}{2g}}$ Where, u DENOTES initial velocity θ denotes angle of projection g denotes accerlation due to gravity By substituting all the given VALUES in the formula, $\dashrightarrow\sf H = \dfrac{u^2 sin^2 \theta}{2g}$ $\dashrightarrow\sf 50= \dfrac{u^2 sin^2 45 \degree}{2(10)}$ $\dashrightarrow\sf 50= \dfrac{u^2 \bigg( \dfrac{1}{ \sqrt{2}} \bigg) ^{2} }{20}$ $\dashrightarrow\sf 50 \times 20= u^2 \bigg( \dfrac{1}{ \sqrt{2}} \bigg) ^{2}$ $\dashrightarrow\sf 1000= \dfrac{ {u}^{2} }{ 2}$ $\dashrightarrow\sf 2000= {u}^{2}$ $\dashrightarrow\sf u = \sqrt{2000}$ $\dashrightarrow \boxed{\sf u = 44.7 \: {ms}^{ - 1} }$ Thus, the initial velocity of the ball is 44.7 m/s.