Question

the criket ball attains a ma height of 50 m and cross a range of 200m in a ground. find its angle of projection​

Answer 1

Answer:

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\blue{\mathfrak{Given}}}}}}}}}$

Max height of projectile = 50 m

Range = 200m

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\blue{\mathfrak{To \:Find}}}}}}}}}$

Angle of Projection

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\blue{\mathfrak{Formulas \: Used}}}}}}}}}$

Max height = H = u² sin²(θ)/2g

Range = R = u² sin(2θ)/g

=> R = u² × 2sin(θ)cos(θ)/g

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\blue{\mathfrak{Calculation}}}}}}}}}$

H = 50 m, R = 200 m

R/H = {2sin(θ)cos(θ)}/{sin²(θ)/2}

=> 200/50 = 4/tan(θ)

=> tan(θ) = 1

=> θ = 45°

$\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\blue{\mathfrak{Final \: Answer}}}}}}}}}$

Angle of projection is 45°.

Answer 2

$\huge\text{\underline{Correct\:Question:-}}$

The cricket ball attains a maximum height of 50 m and cross a range of 200 m in a ground. Find it's angle of projection.

$\bf{\huge{\underline{\boxed{\sf\purple{Angle\:of\:projection\:=45°}}}}}$

$\rule{100}2$

$\textbf{\underline{\underline{Explanation:-}}}$

It is given that, the cricket ball is thrown, it attains maximum height of 50 m and crosses a range of 200 m in  ground.

So,

→ Maximum height = 50 m

→ Range = 200 m.

Also,

→ Maximum height = u² sin² ( θ ) / 2g

→ Range = u² 2 sin ( θ ) cos (θ) / g

∵ Area of projection  $= \dfrac{R}{H}$

Now,

Substitute the given values in the formula.

 {u² 2 sin ( θ ) cos (θ) /g} ÷ {u² sin² ( θ ) / 2g  = $\dfrac{200}{50}$

⇒ 4 / tan ( θ ) = 200 / 50

⇒ tan ( θ ) = 1

∴ cos = 45°

Answer

It's angle of projection is 45°

$\rule{200}2$