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Q.Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2024Subjective· 4mImportance★★★★★
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Writing the horizontal and vertical motions and eliminating time t gives y = (tan theta) x - [g/(2 u^2 cos^2 theta)] x^2, an equation of the form y = ax - bx^2, which is a parabola.

Consider an object projected from the origin with initial speed u at an angle theta above the horizontal. Resolve the initial velocity:

  • Horizontal component: u_x = u cos(theta)
  • Vertical component: u_y = u sin(theta)

The horizontal motion has no acceleration (ignoring air resistance), while the vertical motion has downward acceleration g.

Horizontal displacement after time t:

x = (u cos theta) t ... (1)

Vertical displacement after time t:

y = (u sin theta) t - (1/2) g t^2 ... (2)

From equation (1), express time t:

t = x / (u cos theta) ... (3)

Substitute (3) into (2):

y = (u sin theta) [ x / (u cos theta) ] - (1/2) g [ x / (u cos theta) ]^2

y = (sin theta / cos theta) x - (1/2) g x^2 / (u^2 cos^2 theta)

y = (tan theta) x - [ g / (2 u^2 cos^2 theta) ] x^2

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