Question

x2 is 8 km please help​

Answer 1

Answer:

$v=\frac{800}{\sqrt{3} }  m/s$

Explanation:

Given that,

A body is projected at an angle of 45°(β) with an intial elevation .

We can see that there will be no change in the horizontal component of velocity. Let the velocity of projection be v.

$vcos\beta$ would be the horizontal component.

Now we see that the time taken to decent from the maximum height of $y_{2}$ , the body will cover a horizontal distance of $x_{2}$.

At the maximum height vertical component of velocity is zero so the time taken to fall is given by:-

$vcos\beta *t=x_{2}$

$\beta =45$

$x_{2}=8000m$

$t=\sqrt{\frac{2y_{2} }{g} }$

$t=10\sqrt{6} s$

On the substituting t in abow expression we get

$v=\frac{800}{\sqrt{3} }  m/s$