Question

Aball is thrown from the ground at an angle of 45degree from the horizontal rising to maximum height of 50m find initial velocity

Answer 1

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.