Question

Obtain equation of trojectory of a projectile. ...plzz answer fast​

Answer 1

• ⟹ Path of projectile is given in attachment

Explanation:- As we know that the horizontal component of the volocity will be same cause there is no acceleration on the horizontal axis .

Thus ,the horizontal distance travelled by the projectile is Sₓ = uₓt + ½aₓt²

Here , Sₓ = X , uₓ= ucos¢ , and aₓ = 0

∴ X = ucosθt

⟹ t = X/ucosθ _______(I)

now , The vertical distance travelled by the projectile is Sy = uyt+ ½ayt

∵Sy = Y , uy = usinθ and a = -g

∴Y = usinθ t + ½(-g)t²

Y = usinθt -½gt² _____(II)

But ,from (I)

Y = usinθ × X/ucosθ -½g(x/ucosθ)²

⟹ Y = utanθ- gx²/2u²cos²θ is Equation of trijectory of projectile .

Where ,y = verticle component

X = horizontal component

g = gravity value

u = initial velocity

θ is angle of inclination of initial velocity from horizantal axis .

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Answer 2

Answer:

As we know that the horizontal component of the volocity will be same cause there is no acceleration on the horizontal axis .

Thus ,the horizontal distance travelled by the projectile is Sₓ = uₓt + ½aₓt²

Here , Sₓ = X , uₓ= ucos¢ , and aₓ = 0

∴ X = ucosθt

⟹ t = X/ucosθ _______(I)

now , The vertical distance travelled by the projectile is Sy = uyt+ ½ayt

∵Sy = Y , uy = usinθ and a = -g

∴Y = usinθ t + ½(-g)t²

Y = usinθt -½gt² _____(II)

But ,from (I)

Y = usinθ × X/ucosθ -½g(x/ucosθ)²

⟹ Y = utanθ- gx²/2u²cos²θ is Equation of trijectory of projectile .

Where ,y = verticle component

X = horizontal component

g = gravity value

u = initial velocity

θ is angle of inclination of initial velocity from horizantal axis .

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