The equation of motion of projectile , projected from the top of a tower , is
$y = 12 x - (3 / 4) x^{2}$
The horizontal component of velocity is 3 m/s. What is the range of projectile ? (g = 10 m / s²)

Question

The equation of motion of projectile , projected from the top of a tower , is

$y = 12 x - (3 / 4) x^{2}$

The horizontal component of velocity is 3 m/s. What is the range of projectile ? (g = 10 m / s²)

Infinity Learn · Accepted Answer

Given that, $$y=12x$$−$$(3/4)x2$$∴    $$dydt=12dxdt$$−$$32xdxdt$$ At      $$x=0$$, $$dydt=12dxdtIf$$ angle of projection be θ,  $$then(dy/dt)(dx/dt)=$$$$tan$$⁡θ$$=12If$$ u is initial velocity of projection, then ucos⁡θ$$=3Therefore$$, $$tan$$⁡θ×$$(ucos$$⁡θ$$)=36or$$               usin⁡θ$$=36Now$$,   $$R=u2sin$$⁡$$2$$θ$$g=u2$$×$$2sin$$⁡θ$$cos$$⁡θ$$g=2(usin$$⁡θ$$)(ucos$$⁡θ$$)g=2$$×$$36$$×$$310=21.6cm$$

The equation of motion of projectile , projected from the top of a tower , is
$y = 12 x - (3 / 4) x^{2}$
The horizontal component of velocity is 3 m/s. What is the range of projectile ? (g = 10 m / s²)

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The equation of motion of projectile , projected from the top of a tower , isy=12x−(3/4)x2The horizontal component of velocity is 3 m/s. What is the range of projectile ? (g = 10 m / s2)

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The equation of motion of projectile , projected from the top of a tower , is
$y = 12 x - (3 / 4) x^{2}$
The horizontal component of velocity is 3 m/s. What is the range of projectile ? (g = 10 m / s²)