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Derive the expression for a body projected horizontally. |
Answer» Solution :(i) At any instant t , the projectile has velocity components along both x - axis and y - axis .  (ii) The resultant of these two components gives the velocity of the projectile at that instant t , as shown in Figure (iii) The velocity component at any t along horizontal (x - axis) is `v_(x)=u_(x)+a_(x)t` SINCE, `u_(x)=u, a_(x)=0,`we get `v_(x)=u.` (iv) The component of velocity along vertical direction (y - axis) is `v_(y)=u_(y)+a_(y)t` Since, `u_(y)=0, a_(y)=G,`we get `v_(y)=g t` (V) Hence the velocity of the particle at any instant is `vecv=uhati+g t hatj` (vi) The speed of the particle at any instant t is GIVEN by `THEREFORE""v=sqrt(v_(x)^(2)+v_(y)^(2)), v=sqrt(u^(2)+g^(2)t^(2))` |
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