Question

a bomb at rest at the Summit of a cliff breaks into two equal fragments one of the fragments attains a horizontal velocity 20 root 3 metre per second the horizontal distance between the two fragments when the displacement vectors incline 60° related to each other​

Answer 1

Answer:

Horizontal distance between the two parts is given as

$x = 480\sqrt3$

Explanation:

As we know that the two fragments are of equal mass

So they both will move with same speed in opposite direction

Now we know that the displacement vector makes 60 degree angle with each other

so here we can say that displacement of two parts after "t" time is given as

$d_1 = (20\sqrt3)t \hat i + (-\frac{1]{2}gt^2)\hat j$

and similarly for other part

$d_2 = (-20\sqrt3)t \hat i + (-\frac{1}{2}gt^2)\hat j$

now angle between them is 60 degree

so we have

$d_1. d_2 = d_1 d_2 cos60$

$-1200t^2 + \frac{1}{4}g^2 t^4 = (1200 t^2 + \frac{1}{4}g^2t^4)(0.5)$

$1800 t^2 = \frac{1}{8}g^2t^4$

$t^2 = 18\times 8$

$t = 12 s$

now the horizontal distance between them is given as

$x = 40 \sqrt3 t$

$x = 40\sqrt3 (12)$

$x = 480\sqrt3$

#Learn

Topic : Projectile motion

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