Relative Motion Between two Projectiles
Suppose that two particles are projected from the ground with speeds `u_1` and `u_2` at angles `alpha_1` and `alpha_2` as shown in figure. Acceleration of both the particles is g downwards. So, relative acceleration between them is zero because
`a_(12)=a_1-a_2=g-g=zero`
i.e., the relative motion between the two particles is uniform. Now
`u_(1x) = u_1cosalpha_1`
`u_(2x)= u_2 cosalpha_2`
`u_(1y) = u_1sinalpha_1`
`u_(2y)= u_2 sinalpha_2`
Therefore,
`u_(12x)=u_(1x)-u_(2x)= u_1cosalpha_1- u_2 cosalpha_2`
`u_(12y)=u_(1y)-u_(2y)=u_1sinalpha_1-u_2 sinalpha_2`
`u_(12x)` and `u_(12y)` are the x and y components of relative velocity of 1 with respect to 2.
Hence, relative motion of 1 with respect to 2 is a straight line at an angle `theta=tan^(-1)((u_(12y))/(u_(12x)))`
Now, if `u_(12x) =0` or `u_1 cosalpha_1 = u_2 cosalpha_2`, the relative motion is along y-axis or in vertical direction (as `theta= 90�`).
similarly, if `u_(12y) = 0` or `u_1 sinalpha_1 = u_2 sinalpha_2`, the relative motion is along x-axis or in horizontal direction (as `theta = 0�`).
`text(Condition of Collision of two Projectiles :)`
From the above discussion, it is clear that relative motion between two projectiles is uniform and the path of one projectile as observed by the other is a straight line. Now let the particles are projected simultaneously from two different heights `h_1` and `h_2` with speeds `u_1` and `u_2` in the directions shown in figure. Then the particles collide in air if relative velocity of 1 with respect to 2 `(vecu_(12))` is along line AB or the relative velocity of 2 with respect to 1 `(vecu_(12))` is along the line BA. Thus,
`tan theta=(u_(12y))/(u_(12x))=(h_2-h_1)/s`
Here `u_(12y) = u_1 sinalpha_1 - u_2 sinalpha2`
and `u_(12x) = (u_1 cosalpha_1) - (- u_2 cosalpha_2) = u_1 cosalpha_1 + u_2 cosalpha_2`
If both the particles are initially at the same level `(h_1-h_2)`, then for collision
`u_(12y) = 0` or `u_1 sinalpha_1 = u_2 sinalpha_2`
The time of collision of the two particles will be
`t=(AB)/(|vecu_(12)|)=(AB)/(sqrt((u_(12x))^2 + (u_(12y))^2))`
`a_(12)=a_1-a_2=g-g=zero`
i.e., the relative motion between the two particles is uniform. Now
`u_(1x) = u_1cosalpha_1`
`u_(2x)= u_2 cosalpha_2`
`u_(1y) = u_1sinalpha_1`
`u_(2y)= u_2 sinalpha_2`
Therefore,
`u_(12x)=u_(1x)-u_(2x)= u_1cosalpha_1- u_2 cosalpha_2`
`u_(12y)=u_(1y)-u_(2y)=u_1sinalpha_1-u_2 sinalpha_2`
`u_(12x)` and `u_(12y)` are the x and y components of relative velocity of 1 with respect to 2.
Hence, relative motion of 1 with respect to 2 is a straight line at an angle `theta=tan^(-1)((u_(12y))/(u_(12x)))`
Now, if `u_(12x) =0` or `u_1 cosalpha_1 = u_2 cosalpha_2`, the relative motion is along y-axis or in vertical direction (as `theta= 90�`).
similarly, if `u_(12y) = 0` or `u_1 sinalpha_1 = u_2 sinalpha_2`, the relative motion is along x-axis or in horizontal direction (as `theta = 0�`).
`text(Condition of Collision of two Projectiles :)`
From the above discussion, it is clear that relative motion between two projectiles is uniform and the path of one projectile as observed by the other is a straight line. Now let the particles are projected simultaneously from two different heights `h_1` and `h_2` with speeds `u_1` and `u_2` in the directions shown in figure. Then the particles collide in air if relative velocity of 1 with respect to 2 `(vecu_(12))` is along line AB or the relative velocity of 2 with respect to 1 `(vecu_(12))` is along the line BA. Thus,
`tan theta=(u_(12y))/(u_(12x))=(h_2-h_1)/s`
Here `u_(12y) = u_1 sinalpha_1 - u_2 sinalpha2`
and `u_(12x) = (u_1 cosalpha_1) - (- u_2 cosalpha_2) = u_1 cosalpha_1 + u_2 cosalpha_2`
If both the particles are initially at the same level `(h_1-h_2)`, then for collision
`u_(12y) = 0` or `u_1 sinalpha_1 = u_2 sinalpha_2`
The time of collision of the two particles will be
`t=(AB)/(|vecu_(12)|)=(AB)/(sqrt((u_(12x))^2 + (u_(12y))^2))`