A ball of mass M is thrown vertically upwards. Another ball of mass 2M is thrown at an angle `theta` with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1: cos theta`

Answer

To solve the problem, we need to find the ratio of the heights attained by two balls thrown under different conditions. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have two balls:
- Ball 1 (mass M) is thrown vertically upwards.
- Ball 2 (mass 2M) is thrown at an angle θ with the vertical.

Both balls stay in the air for the same period of time.

### Step 2: Establish Time of Flight
Let \( T_1 \) be the time of flight for Ball 1 and \( T_2 \) for Ball 2. Since both balls stay in the air for the same time, we have:
\[ T_1 = T_2 \]

### Step 3: Time of Flight for Each Ball
For Ball 1 (thrown vertically upwards):
- The time of ascent to the maximum height is given by:
\[ T_1 = \frac{2U_1}{g} \]
where \( U_1 \) is the initial velocity of Ball 1 and \( g \) is the acceleration due to gravity.

For Ball 2 (thrown at an angle θ):
- The vertical component of the initial velocity is \( U_2 \cos \theta \).
- The time of flight is given by:
\[ T_2 = \frac{2U_2 \cos \theta}{g} \]

### Step 4: Set the Times Equal
Since \( T_1 = T_2 \):
\[ \frac{2U_1}{g} = \frac{2U_2 \cos \theta}{g} \]
Cancelling \( 2/g \) from both sides gives:
\[ U_1 = U_2 \cos \theta \]

### Step 5: Calculate the Heights
Now, we calculate the heights attained by both balls.

For Ball 1:
- The maximum height \( h_1 \) is given by:
\[ h_1 = \frac{U_1^2}{2g} \]

For Ball 2:
- The maximum height \( h_2 \) is given by:
\[ h_2 = \frac{(U_2 \cos \theta)^2}{2g} \]
Substituting \( U_1 = U_2 \cos \theta \) into the equation for \( h_1 \):
\[ h_1 = \frac{(U_2 \cos \theta)^2}{2g} \]

### Step 6: Find the Ratio of Heights
Now, we find the ratio of the heights \( h_1 \) and \( h_2 \):
\[ \frac{h_1}{h_2} = \frac{\frac{U_1^2}{2g}}{\frac{(U_2 \cos \theta)^2}{2g}} = \frac{U_1^2}{(U_2 \cos \theta)^2} \]
Substituting \( U_1 = U_2 \cos \theta \):
\[ \frac{h_1}{h_2} = \frac{(U_2 \cos \theta)^2}{(U_2 \cos \theta)^2} = 1 \]

### Conclusion
Thus, the ratio of the heights attained by the two balls is:
\[ h_1 : h_2 = 1 : 1 \]

### Final Answer
The heights attained by the two balls are in the ratio **1:1**.

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