Physics, asked by prakriti16, 1 year ago

a projectile is given an initial velocity of (i+2j) m/s where I is along the ground and j is along vertical. if g=10m/s^2 the equation of trajectory is?

Answers

Answered by abhi178
74

A particle is projected with velocity of (i + j) m/s.

so, horizontal component of velocity, u_x=1m/s

vertical component of velocity, u_y=2m/s

using formula for horizontal component,

x=u_xt+\frac{1}{2}a_xt^2

here, u_x=1,a_x=0

then, x = t ....(1)

applying formula for vertical component,

y=u_yt+\frac{1}{2}a_yt^2

here, u_y=2,a_y=-g

so,y=2t-\frac{1}{2}gt^2

from equation (1),

y=2x-\frac{1}{2}gx^2 this is the required equation of trajectory. here it is clear that equation is not other than parabola.

Answered by AjDevgn
53

Answer:

y=2x-5x²

Explanation:

Initial velocity V= i+2jm/s=Ux i + Uy j=ucos0 i+ usin90 j

then:

tan●=usin90/ucos0=2/1=2

Now:

equation of trajectory:

y = xtan● - 1/2( gx²/u²cos²90)

=> y= 2x-1/2[10x²/(ucos90)²]

=>y=2x-1/2[10x²/(1)²]

=>y=2x-5x²_________ans

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Aur naam yaad rkhna Sultaan Mirza

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