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.