Respuesta :
The velocity with which the particle Q is fired is 15m/s upwards.
What is velocity?
The vector quantity velocity (v), denoted by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).
Assume the upward direction is positive and the downward direction is negative.
Velocity P, = 40m/s
Distance traveled by P,
Using the first equation of motion for particle P,
v = u + at
⇒ 0 = 40 + (-10)t
⇒ t = 4s
This is the time it takes for P to rise
Now, the maximum height(s) reached by the particle P is,
Using the second equation of motion,
s = ut + 1/2at²
⇒ s = 40×4 + 1/2 × (-10) × 4²
⇒ s = 80m
a) A particle P falls when Q is shot after 4 seconds from the initial time:
V² = U² + 2aH₁
⇒ H₁ = 15² - 0/ 2(-10)
⇒ H₁ = 11.25m
Particles P and Q meet at a distance from the ground (H₂).
Height, H₂ = s - H
⇒ H₂ = 80 - 11.25
= 68.75m
Particles P and Q meet at a distance of 68.75m from the ground.
v = u + at₁
⇒ 15 = 0 + (-10) t₁
⇒ t₁ = 1.5s
It is equal to P's fall time and Q's rise time.
b) For particle Q
H₂ = u₂t₁ + 1/2at₁
⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5
⇒ u₂ = 15 m/s
Therefore,
The velocity with which the particle Q is fired is 15m/s upwards.
To learn more about Free Fall under gravity, refer to:
https://brainly.com/question/13299152
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