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To find the time at which both bodies reach the same height, we can analyze their respective motions.

For the body dropped from a height of 300m, we can use the equation of motion:

h₁ = (1/2)gt₁²

Where:

  • h₁ is the height of the dropped body.
  • g is the acceleration due to gravity (approximately 9.8 m/s²).
  • t₁ is the time it takes for the dropped body to reach the desired height.

For the body projected upward at 150m/s, we can use the equation of motion:

h₂ = u₂t₂ - (1/2)gt₂²

Where:

  • h₂ is the height of the projected body.
  • u₂ is the initial velocity of the projected body (150 m/s).
  • t₂ is the time it takes for the projected body to reach the desired height.

We want to find the time at which both bodies reach the same height, so we set h₁ = h₂ and solve for t₂:

(1/2)gt₁² = u₂t₂ - (1/2)gt₂²

Rearranging the equation:

(1/2)gt₁² + (1/2)gt₂² = u₂t₂

Since both bodies start at the same time, t₁ = t₂ = t. Substituting this into the equation:

(1/2)gt² + (1/2)gt² = u₂t

Simplifying the equation:

gt² = 2u₂t

gt = 2u₂

Now, substituting the values g ≈ 9.8 m/s² and u₂ = 150 m/s:

9.8t = 2 * 150

Solving for t:

t = 2 * 150 / 9.8

t ≈ 30.61 seconds

Therefore, both bodies will reach the same height after approximately 30.61 seconds.

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