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

Question

3 years ago

5

ball will reach?

1 answer:

aksik [14] · Answer 1 · 2022-04-01T00:01:18+03:00

aksik [14]3 years ago

8 0

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

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

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

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

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

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

☯ As we know that,

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

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

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

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

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

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

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