A projectile is projected with velocity of 25 m / s at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be: [use g=10 m / s 2 ]

Question

A projectile is projected with velocity of 25 m / s at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be: [use g=10 m / s 2 ] (A) (1/2) sin -1((5 t 2/4 R )) (B) (1/2) sin -1((4 R /5 t 2)) (C)  tan -1((4 t 2/5 R )) (D)  cot -1((R/20 t2))

Tardigrade Admin · Accepted Answer

R =( V 2(2 sin θ cos θ)/ g ) t =( V sin θ/ g ) ⇒ V =( gt / sin θ) ⇒ R =( g 2 t 2/ sin 2 θ) ⋅ (2 sin θ cos θ/ g ) tan θ=(2 gt 2/ R )=(20 t 2/ R ) cot θ=( R /20 t 2)

Q. A projectile is projected with velocity of $25 m / s$ at an angle $θ$ with the horizontal. After $t$ seconds its inclination with horizontal becomes zero. If $R$ represents horizontal range of the projectile, the value of $θ$ will be : [use $g = 10 m / s^{2}$ ]

$\frac{1}{2} sin^{- 1} (\frac{5 t ^{2}}{4 R})$

$\frac{1}{2} sin^{- 1} (\frac{4 R}{5 t ^{2}})$

$tan^{- 1} (\frac{4 t ^{2}}{5 R})$

$cot^{- 1} (\frac{R}{20 t ^{2}})$

$R = \frac{V ^{2} ( 2 s i n θ c o s θ )}{g}$
$t = \frac{V s i n θ}{g} \Rightarrow V = \frac{g t}{s i n θ}$
$\Rightarrow R = \frac{g ^{2} t ^{2}}{s i n ^{2} θ} \cdot \frac{2 s i n θ c o s θ}{g}$
$tan θ = \frac{2 g t ^{2}}{R} = \frac{20 t ^{2}}{R}$
$cot θ = \frac{R}{20 t ^{2}}$

Q. A projectile is projected with velocity of 25m/s at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be : [use g=10m/s2 ]

21​sin−1(4R5t2​)

21​sin−1(5t24R​)

tan−1(5R4t2​)

cot−1(20t2R​)