A shell is projected. from a gun with a muzzle velocity, $u$ . The gun is fitted with a trolley car at an angle $θ$ as shown in Figure. If the trolley car is made to move with constant velocity $v$ towards right. find the
a. horizontal range of the shell relative to ground.
b. horizontal range of the shell relative to a person travelling with trolley.

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Solution

The shell will follow parabolic path as seen from the observer travelling with trolley as well as seen from the observer on ground.
a. The velocity of projection of the shell is ${\to v}_{s} = {\to v}_{s c} + {\to v}_{c}$
Substituting ${\to u}_{s c} = u cos θ^i + u sin θ^j$ and ${\to v}_{c} = v^i$ , we have
${\to u}_{s} = (u c o s θ + v)^i + u sin θ^j$
For horizontal range R of the shell, its displacement in horizontal direction can be given as $\to s = R^i, \to a = - g^j$
and ${\to u}_{s} = (u c o s θ + v)^i + u sin θ^j$
Using $\to s = \to u t + \frac{1}{2} \to a t^{2},$ we have
$R^i = (u cos θ + v) t^i + (u t sin θ - \frac{1}{2} g t^{2})^j$
Comparing the coefficient of $^i$ and $^j$ , we obtain
$R = (u cos θ + v) t$ ....(i)
and $u t sin θ - \frac{1}{2} g t^{2} = 0$ ...(ii)
From (ii) , we find $t = \frac{2 u sin θ}{g}$
Or we would have directly used the formula for the time of flight directly to get same result.
Finally. substituting $t = \frac{2 u sin θ}{g}$ in (i), we have
$R \frac{2 u sin θ (u c o s θ + v)}{g}$
b. The horizontal component of the initial velocity of the shell, as seen from trolley, $u_{x} = u c o s θ .$
The vertical component, $u_{y} = u sin θ .$
The time of flight of shell will be same for both observers at trolley as well as ground. Hence, range as seen from trolley is equal to the horizontal displacement in $x$ direction w.r.t trolley.
$R = (u c o s θ) t = u c o s θ \times (\frac{2 u sin θ}{g}) = \frac{2 u^{2} sin θ . c o s θ}{g} = \frac{u^{2} sin 2 θ}{g}$

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