two bodies are thrown vertically upwards from heights of 2.2m&1.05m,with an initial velocity of 10 ms^-1 .the ratio of their time of flight is..

To determine the ratio of the time of flight for two bodies thrown vertically upwards from different heights, we can use the principles of kinematics. The time of flight for an object thrown upwards can be calculated using the formula derived from the equations of motion. Let's break this down step by step.

Understanding the Motion

When an object is thrown upwards, it will rise until it reaches its maximum height, then it will fall back down. The total time of flight can be divided into two parts: the time taken to reach the maximum height and the time taken to return to the original height from which it was thrown.

Key Variables

• Initial height (h): The height from which the object is thrown.
• Initial velocity (u): The speed at which the object is thrown upwards (10 m/s in this case).
• Acceleration due to gravity (g): Approximately 9.81 m/s², acting downwards.
• Final velocity (v): At the maximum height, the final velocity will be 0 m/s.

Calculating Time of Flight

The time taken to reach the maximum height can be calculated using the formula:

t_{up} = (v - u) / -g

Since the final velocity at the maximum height is 0, we can rearrange this to:

t_{up} = u / g

For the downward journey, the time taken to fall from the maximum height back to the original height can be calculated using the formula:

t_{down} = \sqrt{(2h/g)}

Thus, the total time of flight (T) for each body can be expressed as:

T = t_{up} + t_{down}

Calculating for Each Body

Let's calculate the time of flight for both bodies:

Body 1: Thrown from 2.2 m

For the first body:

• Initial height (h1) = 2.2 m
• Time to reach maximum height (t_{up1}) = 10 m/s / 9.81 m/s² ≈ 1.02 s
• Time to fall back down (t_{down1}) = √(2 * 2.2 m / 9.81 m/s²) ≈ 0.67 s

Thus, the total time of flight for the first body:

T1 = t_{up1} + t_{down1} ≈ 1.02 s + 0.67 s ≈ 1.69 s

Body 2: Thrown from 1.05 m

For the second body:

• Initial height (h2) = 1.05 m
• Time to reach maximum height (t_{up2}) = 10 m/s / 9.81 m/s² ≈ 1.02 s
• Time to fall back down (t_{down2}) = √(2 * 1.05 m / 9.81 m/s²) ≈ 0.46 s

Thus, the total time of flight for the second body:

T2 = t_{up2} + t_{down2} ≈ 1.02 s + 0.46 s ≈ 1.48 s

Finding the Ratio of Time of Flight

Now, we can find the ratio of the time of flight of the two bodies:

Ratio = T1 / T2 ≈ 1.69 s / 1.48 s ≈ 1.14

Final Thoughts

The ratio of the time of flight for the two bodies thrown from heights of 2.2 m and 1.05 m, both with an initial velocity of 10 m/s, is approximately 1.14. This means that the first body takes about 14% longer to complete its flight compared to the second body, primarily due to the greater height from which it was thrown.