A shell is projected from a gun with muzzle velocity u. The gun is fitted with a trolley car at an angle θ as shown in the fig. If the trolley car is made to move with constant velocity v towards right, find the
(i) horizontal range of the shell relative to ground.
(ii) horizontal range of the shell relative to a person travelling with trolley.
A shell is projected from a gun with muzzle velocity u. The gun is fitted with a trolley car at an angle θ as shown in the fig. If the trolley car is made to move with constant velocity v towards right, find the
(i) horizontal range of the shell relative to ground.
(ii) horizontal range of the shell relative to a person travelling with trolley.
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The correct option is C ,
The shell will follow parabolic path as see from an observer travelling with trolley as well as seen from observer on ground.
(i) The velocity of projection of the shell is →vs = −→vsc + →vc. Substituting −→usc = u cos θ ^j and →vc = v^i
For horizontal range R of the shell its displacement in horizontal direction can be given as →s = R^i
→a = −g^j & →us = (u cos θ + v)^i + u sin θ ^j
Using →s = →ut + 12→at2
We have R^i = (u cos θ + v)t^i + (ut sinθ − 12gt2)^j
Comparing the coefficient of ^i and ^j,
We obtain R = (u cos θ + v)t ..............(i) and ut sin θ − 12gt2 = 0............(ii)
From equation (ii), we find t = 2u sin θg
or we would have directly used the formula for time of flight directly to get same result.
Finally, substituting t = 2u sin θg in equation (i), we have R = 2u sin θ(u cosθ + v)g
(ii) Horizontal component of initial velocity shell as seen from trolley ux = u cosθ
The vertical component uy = u sin θ
Time of flight of shell will be same for both observers at trolley as well as ground.
Hence range as seen from trolley = horizontal displacement in x direction w.r.t. trolley.
R′ = (u cosθ)t = u cos θ × (2u sin θg) = 2u2 sinθ . cosθg = u2 sin 2θg
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The shell will follow parabolic path as see from an observer travelling with trolley as well as seen from observer on ground.
(i) The velocity of projection of the shell is →vs = −→vsc + →vc. Substituting −→usc = u cos θ ^j and →vc = v^i
For horizontal range R of the shell its displacement in horizontal direction can be given as →s = R^i
→a = −g^j & →us = (u cos θ + v)^i + u sin θ ^j
Using →s = →ut + 12→at2
We have R^i = (u cos θ + v)t^i + (ut sinθ − 12gt2)^j
Comparing the coefficient of ^i and ^j,
We obtain R = (u cos θ + v)t ..............(i) and ut sin θ − 12gt2 = 0............(ii)
From equation (ii), we find t = 2u sin θg
or we would have directly used the formula for time of flight directly to get same result.
Finally, substituting t = 2u sin θg in equation (i), we have R = 2u sin θ(u cosθ + v)g
(ii) Horizontal component of initial velocity shell as seen from trolley ux = u cosθ
The vertical component uy = u sin θ
Time of flight of shell will be same for both observers at trolley as well as ground.
Hence range as seen from trolley = horizontal displacement in x direction w.r.t. trolley.
R′ = (u cosθ)t = u cos θ × (2u sin θg) = 2u2 sinθ . cosθg = u2 sin 2θg
