Question

A cricketer thrown a ball at an angle of 30 degree with the horizontal having velocity 30 m/s the maximum heigh attained by the ball is take gravity 10m/s​

Answer 1

Given :

• Angle of projection made with the horizontal, θ = 30°
• Initial velocity of Ball, u = 30 m/s
• Acceleration of gravity to be taken, g = 10 m/s²

To find :

• Maximum height attained by ball, H =?

Formula required :

• Formula to calculate maximum height reached by the projectile

$\red{\bigstar}\boxed{\sf{H=\dfrac{u^2\;sin^2\theta}{2\;g}}}$

[Where H is the maximum height attained by projectile, when thrown with an initial velocity of u, at an angle of projection θ and g is acceleration due to gravity]

Solution :

Using formula for maximum height of attained by projectile

$\implies\sf{H=\dfrac{u^2\;sin^2\theta}{2\;g}}$

$\implies\sf{H=\dfrac{(30)^2\;sin^2 30^{\circ}}{2\;(10)}}$

$\implies\sf{H=\dfrac{900\times sin^230^{\circ}}{20}}$

$\implies\sf{H=45\times sin^2 30^{\circ}}$

$\implies\sf{H=45\times \left(\dfrac{1}{2}\right)^2}$

$\implies\sf{H=45\times \dfrac{1}{4}}$

$\implies\boxed{\boxed{\red{\sf{H=11.25\;\;\;m}}}}\;\;\;\red{\bigstar}$

Therefore,

• Maximum height attained by the Projectile is 11.25 metres.