The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by y = $(4 t - 2 t^{2})$ m and x = (3t) metre, where t is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is

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Solution

Equation of trajectory is $Y = X tan θ - \frac{g X^{2} (1 + {tan}^{2} θ)}{2 u^{2}}$
On Comparison with given equation
$x tan θ = 4 t$
$\Rightarrow (3 t) tan θ = 4 t$
$\Rightarrow tan θ = \frac{4}{3}$

$θ = {tan}^{- 1} (4 / 3) \approx 53^{0}$

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Similar questions

Q. The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by

y = (4 t - 2 t^{2}) m

and

x = (3 t) m

, where

t

is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is

Q. The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by

y = (4 t - 2 t^{2}) m

and

x = (3 t) m

, where

t

is in seconds and point of projection is taken as origin. The angle of projection of projectile with vertical is

The position coordinates of a projectile projected from the ground on a certain planet (with no atmosphere) are given by $y = (4 t - 2 t^{2}) m$ and $x = (3 t)$ meter, where t is in second and point of projection is taken as the origin. The angle of projection of projectile with vertical is

Q. The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by

y = (4 t - 2 t^{2}) m

and

x = (3 t) m

, where

t

is in seconds and point of projection is taken as origin. The angle of projection of projectile with vertical is

Q. The vertical height

y

and horizontal distance

x

of a projectile on a certain planet are given by

x = (3 t) m, y = (4 t) m

where

t

is in seconds. Find the speed of projection

(i n m / s)

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The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by y = (4t−2t2) m and x = (3t) metre, where t is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is

Equation of trajectory is Y=Xtanθ−gX2(1+tan2θ)2u2On Comparison with given equation xtanθ=4t⇒(3t)tanθ=4t⇒tanθ=43θ=tan−1(4/3)≈530

The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by y = $(4 t - 2 t^{2})$ m and x = (3t) metre, where t is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is

Equation of trajectory is $Y = X tan θ - \frac{g X^{2} (1 + {tan}^{2} θ)}{2 u^{2}}$
On Comparison with given equation
$x tan θ = 4 t$
$\Rightarrow (3 t) tan θ = 4 t$
$\Rightarrow tan θ = \frac{4}{3}$

$θ = {tan}^{- 1} (4 / 3) \approx 53^{0}$