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Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle theta with respect to the horizontal direction. |
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Answer» SOLUTION :(i) Consider an OBJECT thrown with initial VELOCITY `vecu` at an angle `theta` with the horizontal. (ii) Since ACCELERATION due to gravity acts vertically downwards, velocity along the horizontal x - direction `u_(x)` doesn't change throught the motion. Whereas velocity along the y-direction `u_(y)` is changed.  The path of the projectile : (a) Motion along x - direction : (i) The horizontal distance travelled by the projectile at a point P after a time t can be written as, `s_(x)=u_(x)t+(1)/(2)a_(x)^(2)` (ii) Here, `s_(x)=x, u_(x)=u cos theta, and a_(x)=0,` Therefore, `x=u cos.t` `t=(x)/(u cos theta)"....(1)"` (b) Motion along y - direction : (i) The downward distance travelled by the projectile at a point P after a time t can be written as, `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)` (ii) Here, `s_(y)=y, u_(y)=u sin theta, and a_(y)=-g` Therefore, `y=u sin theta t-(1)/(2g t^(2)` (iii) Substituting equation (1) , weget, `y=usin theta.(x)/(ucos theta)-(1)/(g)g((x)/(ucos theta))^(2)` `y=x tan theta -(1)/(2)g(x^(2))/(u^(2)cos^(2)theta)` (iv) Thus, the path travelled by the projectile is an inverted parabola. MaximumHeight : `(h_("max"))` (i) The maximum vertical distance travelled by the projectile during its JOURNEY is called maximum height. (ii) For the vertical part of the motion, `v_(y)^(2)=u_(y)^(2)+2a_(y)s_(y)` (iii) Here, `v_(y)=0, s_(y)=h_("max"), u_(y)=u sin theta and a_(y)=-g` Therefore. `0=u^(2)sin^(2)theta-2gh_("max")` `h_("max")=(u^(2)sin^(2)theta)/(2g)` Time of flight : `(T_(f))` (i) The time of flight `(T_(f))` is the total time taken by the projectile to hit the ground after thrown. (ii) The downward distance travelled by the projectile at a time t can be written as, `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)` (iii) Here substituting the values `S_(y)=0, t=T_(f)` `u_(f)=u sin theta, and a_(y)=-g` we get, `0=u sin thetaT_(f)-(1)/(2)gT_(f)^(2)` Therefore,`T_(f)=(2usintheta)/(g)` Horizontial range : (R) (i) The horizontal range (R) is the maximum horizontal distance between the point of projection and the point where the projectile hits the ground. (ii) The horizontal distance travelled by the projectile at a time t can be written as, `s_(x)=u_(x)t+(1)/(2)a_(x)t^(2)` (iii) Here, `S_(x)=R, u_(x)=u cos theta, a_(x)=0 and t=T_(f)` `R=u cos.T_(f)` `R=u cos theta.((2usin theta)/(g))=(2u^(2)sin theta cos theta)/(g)` `""[because T_(f)=(2u sin theta)/(g)]` (iv) Therefore,`""R=(u^(2)sin 2theta)/(g)` `[because 2 theta=2sin theta.cos theta]` (v) For maximum range, `sin 2theta=1` `2theta=(pi)/(2)` `theta=(pi)/(4)` The maximum range is, `R=(u^(2))/(g)` |
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