Question

A body is projected vertically upward with 50 ms⁻¹, it crosses a point above the ground at t = 3 s. Find the time interval (in sec) from this instant after which the body passes through the same point during the return journey.

Answer 1

4

Answer 2

The problem describes a body projected vertically upward and asks for the time interval between its upward and downward passage through a specific point.

1. Understand the Motion: When a body is projected vertically upward, its motion is symmetric.

• The time taken to reach the maximum height is t_peak = u/g.
• The time taken to fall from the maximum height back to the ground is also t_peak.
• For any given height below the maximum, the speed of the body while going up is equal to its speed while coming down.
• The time taken to reach a certain height during the upward journey and the time taken to fall back to that same height from the peak are symmetric.

2. Calculate Time to Reach Maximum Height: Given initial velocity, u = 50 ms⁻¹. Assuming acceleration due to gravity, g = 10 ms⁻² (a common approximation in such problems unless specified otherwise). The time to reach maximum height (t_peak) is given by: t_peak = u / g t_peak = 50 / 10 = 5 s

3. Apply Symmetry for the Given Point: The body crosses a point above the ground at t₁ = 3 s during its upward journey.

• Time taken from the initial projection to reach the peak: t_peak = 5 s.
• Time taken to travel from the point (t₁ = 3 s) to the maximum height: Δt_up = t_peak - t₁ = 5 s - 3 s = 2 s.

Due to symmetry, the time taken for the body to fall from the maximum height back to the same point will be equal to Δt_up.

• Time taken to travel from the maximum height back to the point: Δt_down = 2 s.

4. Calculate the Time of Return Journey and the Interval: The total time elapsed from the start until the body passes through the same point during its return journey (t₂) will be: t₂ = t_peak + Δt_down t₂ = 5 s + 2 s = 7 s

The question asks for the time interval from the instant t = 3 s (upward journey) after which the body passes through the same point during the return journey. This interval is t₂ - t₁. Time interval = t₂ - t₁ = 7 s - 3 s = 4 s.

Alternative Method (using the formula from the similar question): The time interval from the upward passage through a point (at time t₁) to the downward passage through the same point is given by 2 * (u/g - t₁). Time interval = 2 * (50/10 - 3) Time interval = 2 * (5 - 3) Time interval = 2 * 2 Time interval = 4 s

The final answer is $\boxed{\text{4}}$.