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

(a) The position where particle Q and P meet is 68.75 m.

(b) The velocity with which particle Q is shot is 15 m/s.

Time at which the two particles meet each other

The time elapsed before the two particles meet is calculated as follows;

Distance Q - Distance P = Distance between them

Distance traveled by particle P:

Trise = (V - V₀)/g

Trise = (0 - 40) / -10

Trise = 4.0 s

Hmax = V₀t + 0.5gt²

Hmax = 40 x 4 - (0.5)(10)4²

Hmax = 80 m.

P is falling when Q is moving up:

h = Hmax - (V² - V₀²)/2g

h = 80 - ((15)²- 0) / 20

h = 68.75 m.

Thus, the position where particle Q and P meet is 68.75 m

Tfall = (V - V₀)/g = (15-0) / 10

Tfall = 1.5 s

Fall time. = Rise time  for Q.

Distance traveled by particle Q:

h = V₀t + 0.5gt²

h = 80 - 68.75

h = 11.25

V₀ x 1.5 - 5(1.5)² = 11.25

V₀ x 1.5 - 11.25 = 11.25

V₀ x 1.5 = 22.5

V₀ = 22.5 / 1.5

V₀ = 15 m/s

Thus, the velocity with which particle Q is shot is 15 m/s.

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