Final answer:
The two particles meet after 2.5 seconds and Particle Q was shot up with a velocity of 25m/s.
Step-by-step explanation:
In projectile motion, the vertical and horizontal motions can be treated separately. Let's first calculate the time it takes for both particles to meet. Particle P was shot up with a velocity of 40m/s and after 4s, Particle Q was also shot up. If at the point of meeting, P has a velocity of 15m/s, we can use the formula Vfinal = Vinitial + (-g)t to find the time it took for Particle P to reach that point. Rearranging the formula, we get -g*t = Vfinal - Vinitial, and plugging in the given values, -10*t = 15 - 40. Solving for t, we find that t = 2.5s. Therefore, both particles meet after 2.5s.
To find the velocity with which Particle Q was shot up, we need to calculate the initial velocity of Particle Q using the formula y = Vinitial*t + (1/2)(-g)t^2, where y is the displacement and t is the time it took Particle Q to reach the meeting point. Rearranging the formula, we get Vinitial = (y - (1/2)(-g)t^2) / t. Plugging in the given values, Vinitial = 0 - (1/2)(-10)(2.5^2) / 2.5. Simplifying the equation, we find that Vinitial = 25m/s. Therefore, Particle Q was shot up with a velocity of 25m/s.