A particle is projected in such a way that, taking point of projection as origin, y = 8t - 5t2 and x = 6t. What is the range of projectile?

Question

A particle is projected in such a way that, taking point of projection as origin, y = 8t - 5t2 and x = 6t. What is the range of projectile? (A) 48 m (B) 4.8 m (C) 9.6 m (D) 24 m

Tardigrade Admin · Accepted Answer

For range y=0 ⇒ t=0, (8/5) R=6((8/5))=9.6 m

Q. A particle is projected in such a way that, taking point of projection as origin, $y = 8 t - 5 t^{2}$ and $x = 6 t$ . What is the range of projectile?

$48 m$

$4.8 m$

$9.6 m$

$24 m$

For range
$y = 0 \Rightarrow t = 0, \frac{8}{5}$
$R = 6 (\frac{8}{5}) = 9.6 m$

Q. A particle is projected in such a way that, taking point of projection as origin, y=8t−5t2 and x=6t. What is the range of projectile?

48m

4.8m

9.6m

24m