Question

A rocket moves horizontally with a constant velocity $u$ at a height $h$ . A guided missile is fired vertically with a speed when the rocket passes above it.

Assuming that the missile always aims at the rocket with the constant speed $v$ , find the time after the missile strikes the rocket.

Accepted Answer

Consider at any instant the velocity vector of missile makes an angle $θ$ with the velocity vector of the rocket and after time $t$ missile strikes the rocket.

The approach velocity missile w.r.t. rocket can be written as

Velocity of $A$ w.r.t. $B$ at given instant $(v - u \cos θ)$

If $r$ is the distance of separation between rocket and missile after a time $t .$
$(v - u \cos θ) = - \frac{d r}{d t}$
$(v - u \cos θ) d t = - d r$

Integrating the above equation on both sides,

From $t = 0$ to $t = T$ and $r = h$ to $r = 0$ , We get

$\int_{0}^{T} v d t - \int_{0}^{T} u \cos θ d t = - \int_{h}^{0} d r$

$v T - u \int_{0}^{T} \cos θ d t = h . . . . . . . (1)$

The missile will strike the rocket when the displacement in the horizontal direction for missile and rocket is equal.

The velocity of missile in horizontal direction is $v \cos θ$ and for rocket the horizontal velocity is $u$ .

After time $d t$ both missile and rocket will cover the same displacement in the horizontal direction . So, we can write for time $t = 0$ to $t = T$

$\int_{0}^{T} v \cos θ d t = \int_{0}^{T} u d t v \int_{0}^{T} \cos θ d t = u T$
$\int_{0}^{T} \cos θ d t = \frac{u T}{v} . . . . . . . (2)$

From equation $(1)$ and $(2)$ ,We can write

$v T - u (\frac{u T}{v}) = h T = \frac{h v}{v^{2} - u^{2}}$

After time $T = \frac{h v}{v^{2} - u^{2}}$ missile will hit the rocket.

B M Sharma Solutions for Chapter: Kinematics II, Exercise 75: ILLUSTRATION 5.71

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 75: ILLUSTRATION 5.71

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 75: ILLUSTRATION 5.71 with Hints & Solutions

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