Question

A particle is projected from a tower as shown in the figure, then find the distance from the foot of the tower where it will strike the ground.

• A. $1500/3m$
• B. $2500/3m$
• C. $4000/3m$
• D. $2000/3m$

Answer 1

The correct option is C $4000/3m$

We have,

$h=-1500m$

$v=\frac{500}{3}m/s$

So,By using $s=ut+\frac{1}{2}a{t}^2$

$–1500=-100t–\frac{1}{2}\times 10t^2$

$t^2+10t–150=0$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow t=\frac{(-20\pm 40)}{2}$

So, $t=10s$ It means it is the horizontal distance,

Thus,

$d=\frac{500}{3}\times 0.8\times 10$

$=\frac{4000}{3}m$