Question

A ball is throw at an angle of 30 degrees off the horizontal, with an initial velocity of 28 m/s. what is the maximum height the ball will reach?​

Answer 1

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}}$

☯ As we know that,

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

$\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }$

$\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }$

$\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}$

$\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }$

$\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }$

$\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}$