Question

A particle is projected in such a way that,taking point of projection as origin,y=8t-6t^2 and x=6t.what is the range of projectile?

Answer 1

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Answer 2

Answer:

The range of projectile is 8 meter.

Explanation:

Given,

          $y = 8t - 6t^2$  .................................................(i)

  and  $x =6t$         ...................................................(ii)

Equation of a trajectory in projectile motion ,

    $y= u\sin\theta \ t - \frac{1}{2}gt^2$    ................................................(iii)

   $x = u\cos\theta \ t$ ...............................................................(iv)

Comparing eqn. (i) and (iii),we can write

  $u\sin\theta = 8$      and  $\frac{g}{2} = 6$   → $g =12$ $m/sec^2$

Comparing eqn.(ii) and (iv),we will get

               $u\cos\theta = 6$

Horizontal range, $R = \frac{u^2\sin2\theta}{g}$ $= \frac{2u\sin\theta (u \cos\theta)}{g}$

    ∴$R = \frac{2* 8 *6}{12}$  = 8 meter