Show that the projectile angle - for projectie launched from the origin is given by-tan -1-4 H - R -

H = v_0^2 sin^2 θ/2g R = v_0^2 sin 2 θ/g = v_0^2 sin θ cos θ/g 4H/R = ( 4v_0^2 sin^2 θ/2g )× ( g/2 v_0^2 sin θ cos θ ) = tan θ Thus, θ =tan^ 1 [ 4H/R ] ...

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Standard XII

Mathematics

Properties of Determinants

Show that the...

Question

Show that the projectile angle $θ$ for projectie launched from the origin is given by:
$θ = t a n^{- 1} [\frac{4 H}{R}]$

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Solution

$H = \frac{v_{0}^{2} s i n^{2} θ}{2 g} R = \frac{v_{0}^{2} s i n 2 θ}{g} = \frac{v_{0}^{2} s i n θ c o s θ}{g} \frac{4 H}{R} = (\frac{4 v_{0}^{2} s i n^{2} θ}{2 g}) \times (\frac{g}{2 v_{0}^{2} s i n θ c o s θ}) = t a n θ$
Thus, $θ = t a n^{- 1} [\frac{4 H}{R}]$

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Similar questions

Q. Show that the projection angle

0^{o}

for a projectile launched origin is given by

θ_{0} = {tan}^{- 1} (\frac{4 h m}{R})

where, the symbols have their usual meanings.

(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

(b) Show that the projection angle for a projectile launched from the origin is given by

Where the symbols have their usual meaning.

Q. Show that, (a) for a projectile the angle between the velocity and the x-axis as a function of time is given by

θ (t) = t a n^{- 1} (\frac{v_{o} y - g t}{v_{o} x})

(b) Shows that the projection angle

θ_{o}

for a projectile launched from the origin is given by

θ_{o} = t a n^{- 1} (\frac{4 h_{m}}{R})

where the symbols have their usual meaning.

Q. The position of a projectile launched from the origin at

t = 0

is given by

\to r = (40^i + 50^j) m

t = 2 s

. If the projectile was launched at an angle

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(Take g = 10 {ms}^{- 2})

Q. The position of a projectile launched from the origin at

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\to r

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Show that the projectile angle θ for projectie launched from the origin is given by: θ=tan−1[4HR]

H=v20sin2θ2gR=v20sin2θg=v20sinθcosθg4HR=(4v20sin2θ2g)×(g2v20sinθcosθ)=tan θ Thus, θ=tan−1[4HR]

Show that the projectile angle $θ$ for projectie launched from the origin is given by:
$θ = t a n^{- 1} [\frac{4 H}{R}]$

$H = \frac{v_{0}^{2} s i n^{2} θ}{2 g} R = \frac{v_{0}^{2} s i n 2 θ}{g} = \frac{v_{0}^{2} s i n θ c o s θ}{g} \frac{4 H}{R} = (\frac{4 v_{0}^{2} s i n^{2} θ}{2 g}) \times (\frac{g}{2 v_{0}^{2} s i n θ c o s θ}) = t a n θ$
Thus, $θ = t a n^{- 1} [\frac{4 H}{R}]$