A projectile is given an initial velocity of ( hati + 2 hatj) m/s, where hati is along the ground and hatj is along the vertical. If g = 10 m/s2, then the equation of its trajectory is

Question

A projectile is given an initial velocity of ( hati + 2 hatj) m/s, where hati is along the ground and hatj is along the vertical. If g = 10 m/s2, then the equation of its trajectory is (A) y=x-5x2 (B) y=2x-5x2 (C) 4y=2x-5x2 (D) 4y=2x-25x2

Tardigrade Admin · Accepted Answer

Initial velocity = (i + 2j)m/s
Magnitude of initial velocity u =

(1)^{2} + (2)^{2} = 5 m / s

Equation of trajectory of projectile is

y = x t an θ - \frac{g x ^{2}}{2 u ^{2}} (1 + t a n^{2} θ) [t an θ = \frac{y}{x} = \frac{2}{1} = 2]

∴ y = x \times 2 - \frac{10 ( x ) ^{2}}{2 ( 5 ) ^{2}} [1 + (2)^{2}]

= 2 x - \frac{10 ( x ^{2} )}{2 \times 5} (1 + 4)

2 x - 5 x^{2}

Q. A projectile is given an initial velocity of $(\hat{i} + 2 \hat{j}) m / s$ , where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 m / s^{2}$ , then the equation of its trajectory is

$y = x - 5 x^{2}$

$y = 2 x - 5 x^{2}$

$4 y = 2 x - 5 x^{2}$

$4 y = 2 x - 25 x^{2}$

Q. A projectile is given an initial velocity of (i^+2j^​)m/s, where i^ is along the ground and j^​ is along the vertical. If g=10m/s2, then the equation of its trajectory is

y=x−5x2

y=2x−5x2

4y=2x−5x2

4y=2x−25x2

Initial velocity = (i + 2j)m/s Magnitude of initial velocity u =(1)2+(2)2​=5​m/s Equation of trajectory of projectile is y=xtanθ−2u2gx2​(1+tan2θ)[tanθ=xy​=12​=2] ∴y=x×2−2(5​)210(x)2​[1+(2)2] =2x−2×510(x2)​(1+4) 2x−5x2

Q. A projectile is given an initial velocity of $(\hat{i} + 2 \hat{j}) m / s$ , where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 m / s^{2}$ , then the equation of its trajectory is

$y = x - 5 x^{2}$

$y = 2 x - 5 x^{2}$

$4 y = 2 x - 5 x^{2}$

$4 y = 2 x - 25 x^{2}$