Question

A ball is themrown such that maximum height is 45m and range is 180m. Find u and theta?
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Answer 1

Question :

A ball is thrown such that maximum height is 45m and 180m. Find u and $\theta.$

Answer :

Given :

R = 180 m

H = 45 m

where -

$\longrightarrow$R is range.

$\longrightarrow$H is maximum height.

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To Find :

Initial veloicty $\longrightarrow$ u

Angle of projection $\longrightarrow$ $\theta$

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Formula used :

$\boxed{\rm\pink{ R = \dfrac{u^2sin2\theta}{g}}}$

$\boxed{\rm\pink{H = \dfrac{u^2sin^2\theta}{2g}}}$

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Solution :

Substituting the value of R and H in the equation -

$\rm 180 = \dfrac{u^2sin2\theta}{g} \longrightarrow eq 1$

$\rm 45 = \dfrac{u^2sin^2\theta}{2g} \longrightarrow eq 2$

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$\rm\red{Calculating\: angle\: of \:projection\: (\theta)}$ -

By dividing equation 1 by 2

$\implies$$\rm \dfrac{180}{45} = \dfrac{\dfrac{ \cancel{u^2}sin2\theta}{\cancel{g}}}{\dfrac{\cancel{u^2}sin^2\theta}{2\cancel{g}}}$

$\implies$$\rm 4 = \dfrac{2sin2\theta}{sin^2\theta}$

$\implies$$\rm 4 sin^2\theta = 2sin2\theta$

$\implies$$\rm 2 sin^2\theta = sin2\theta$

$\rm\purple{ sin2\theta = 2 sin\theta cos\theta}$

$\implies$$\rm 2 sin^2\theta = 2 sin\theta cos\theta$

$\implies$$\rm sin\theta = cos\theta$

$\implies$$\rm \theta = 45\degree$

$\rm\pink{<u>Angle</u><u>\:</u> <u>of</u> <u>\:</u><u>projection</u> = <u>45\degree</u>}$

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$\rm\red{Calculating\: Intitial\: velocity\: (u) }$

Substituting the value of $\rm \theta$ in eq 1

$\implies$$\rm 180 = \dfrac{u^2sin2\theta}{g}$

$\implies$$\rm 180 = \dfrac{u^2sin2 (45) }{g}$

$\implies$$\rm 180 = \dfrac{u^2sin 90 }{g}$

$\implies$$\rm 180 = \dfrac{u^2}{g}$

$\implies$$\rm u^2 = 180 \times 9.8$

$\implies$$\rm u^2 = 1764$

$\implies$$\rm u = 42 m/s$

$\rm\pink{<u>Initial</u> <u>\:</u><u>velocity</u> = <u>42</u> <u>m/s</u>}$

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