Question

2. Two particles P and Q are shot vertically up. T
Particle P is first up with a velocity 40ms-¹. After
4s, particle Q is also shot up. Find 01 -
(a) where the two particles meet, if at the point of
meeting, P has velocity of 15ms™¹.
(b) the velocity with which Q is shot. (g = 10ms-²)
(000)(9)

Answer 1

Free Fall under gravity and time of meeting is used to find out the required answer.

∴ Let us assume, the upward direction to be positive and and downward direction to be negative.

Given, Velocity of P, = 40m/s

Distance travelled by P,

∴Using First Equation of motion for particle P,

v = u + at

⇒ 0 = 40 + (-10)t

⇒ t = 4s  ; which is the time taken by P to rise up

Now, Maximum Height(s) reached by particle P,

∴Using second equation of motion,

s = ut + 1/2at²

⇒ s = 40×4 + 1/2 × (-10) × 4²

⇒ s = 80m

a) Particle P is falling when Q is shot up after 4s from the initial time

∴ Using third equation of motion using Free Fall under Gravity,

V² = U² + 2aH₁

⇒ H₁ = 15² - 0/ 2(-10)

⇒ H₁ = 11.25m

Hence, Particle P and Q meet at a distance (H₂) from ground,

∴ Height, H₂ = s - H

⇒ H₂ = 80 - 11.25 = 68.75m

Hence, Particle P and Q meet at a distance of 68.75m from ground.

∴ Using First equation of motion for Particle P using the Time of Meeting,

v = u + at₁

⇒ 15 = 0 + (-10) t₁

⇒ t₁ = 1.5s  ; which is equal to the fall time of P and Rise time of Q

b) For particle Q

∴ Using second equation of motion,

H₂ = u₂t₁ + 1/2at₁

⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5

⇒ u₂ = 15 m/s

Hence, the Velocity with which Particle Q was shot is 15m/s in the upward direction.

Learn more about Free Fall under gravity here, https://brainly.com/question/13299152

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