When I am trying to solve the question shown in the image below, I find the displacement vectors for each particles. To find the meeting points, I make the $x$-components equal each other. And I find that the particles do not meet. However the given answer says they do at $2$ seconds.
How?
My Displacement Vectors are as follows: \begin{align} P(S) &= 19.2t ~\hat i + (14.4t-4.9t^2)~\hat j \\ Q(S) &= 10.8t ~\hat i + (14.4t-4.9t^2) ~\hat j \end{align}
On
The answer by John is not correct, The time at impact between the two objects assuming $g = -9.8$ will be $300/144$ seconds( Critical the rounding for this type of question is not done until the very end) Also, The final answer of 38 meters is definitely not correct, The answer is $8.7$ Metres above the ground, I suspect he didn't plug in the value of 2 seconds into his equation correctly, as there is no way $t=2, y = 38,$
As per $y= -4.9^2 +14.4t$ When $t = 2,$ $y = 9. 2$ Not $38$ meters above the surface.
Notice that the two particles each have the same initial upward velocity of $14.4$ meters per second. This means that their vertical displacements will be identical. This also means that the only way the two particles will not collide is if they are far enough apart that they hit the ground before coming together.
The expression for the vertical displacement $y$ is $y=-4.9t^2 + 14.4t$ and this equals $0$ when either $t=0$ or $t=\frac{14.4}{4.9}\approx 2.94$ seconds.
Now, $x_A=v_0(\cos\theta) t=19.2t$ and $x_B=v_0(\cos\phi)t + x_0=-10.8t + 60$.
Setting $19.2t=-10.8t+60$ gives us $30t=60\implies t=2$.
This tells us that the particles will collide $38.4$ meters east of $A$ and at an altitude of $38$ meters.