Question

a particle is thrown upwards with a velocity of 40m/s. It attains a height (h) after (t) seconds. another particle is thrown from a height of 100m at the same time. let them meet at t second. find t and h​

Answer 1

Given:-

For First Particle

• Initial Velocity of the Particle = 40m/s

• Acceleration due to gravity = -10m/s²

• Distance Covered = (100 - x)

For Second Particles:-

• Distance Covered = x

• Initial Velocity = 0m/s

• Acceleration due to gravity = +10m/s²

To Find:-

• The time when they meet and the Height the meet.

Formulae used:-

• S = ut + ½ × a × t²

Where,

• S = Distance
• u = Initial Velocity
• a = Acceleration
• t = Time.

Now,

For The second Particles.

→ S = ut + ½ × a × t²

→ x = (0) (t) + ½ × (10) × t²

→ x = 5t²......1

Therefore,

For First Particle

→ S = ut + ½ × a × t²

→ (100 - x) = (40)(t) + ½ × -10 × t²

→ (100 - x) = 40t -5t²

→ 100 - 5t² = 40t - 5t²

→ 100 = 40t - 5t² + 5t²

→ 100 = 40t

→ t = 100/40

→ t = 2.5s

Atq. At the same time the Second particles was also thrown.

So,

→ x = 5t²

→ x = 5 (2.5)²

→ x = 5 × 6.25

→ x = 31.25m

So,

The Second stone will meet at 2.5s at height of 31.25 from the top and (100 - x) → (100 - 31.25) → 68.75. from down