Physics, asked by shireenball966, 9 months ago

Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

Answers

Answered by abdul9838

$<b> <body \: bgcolor = "pink">$

$\bf \orange{hey \: mate \: here \: is \: answer} \\ \: \\ \bf \orange{according \: to \: the \: question} \\ \\ \bf \orange{horizontal \: range(r) = maximum \: height \: (h)} \\ \\ \bf \orange{ \frac{ \cancel{{u}^{2} }\sin(2 \theta) }{ \cancel{g}} = \frac{ \cancel {u}^{2} { \sin }^{2} \theta}{2 \cancel{g}} } \\ \\ \bf \orange{ \sin(2 \theta) = \frac{ \sin( \theta) \times \sin( \theta) }{2} } \\ \\ \bf \orange{2 \cancel{\sin( \theta )} \times \cos( \theta) = \frac{ \cancel{\sin( \theta) } \times \ {sin( \theta) }}{2} } \\ \\ \bf \orange{2 \cos( \theta) = \frac{ \sin( \theta) }{2} } \\ \\ \bf \orange{ \frac{ \sin( \theta) }{ \cos(\theta)} = 4} \\ \\ \bf \orange{tan \theta = 4} \\ \\ \bf \orange{ {tan}^{ - 4} } \\ \\ \bf \orange{or} \\ \\ \bf \orange{ \theta = 76 \degree}$

Answered by techtro

The angle of projection of a projectile for which the horizontal range and maximum height are equal is :

• Given : Horizontal range = maximum height

• We know, horizontal range is given by formula

R = u^2×sin2θ/g

And maximum height is given by,

H = u^2×sin^2θ/2g

• When R = H,

u^2×sin2θ/g = u^2×sin^2θ/2g

sin2θ = sin^2(θ) / 2

2sinθ×cosθ = sinθ×sinθ/2

2cosθ = sinθ/2

tanθ = 4

• θ = tan^-1(4) = 76°

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