Question

A ball of mass m is thrown vertically upward. Another ball of mass 2 m is thrown at an angle θ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is 1/x. The value of x is _______.

Answer 1

Correct Answer: 1

To solve this problem, we need to compare the heights attained by two balls using their respective flight properties. When the ball is thrown vertically, its equations of motion determine its maximum height. Meanwhile, for the ball thrown at an angle, the vertical component of its velocity determines its time in the air and maximum height. The problem states both balls are in air for the same time. We will use this information to find the ratio of heights. 

Let:
- \( m \) be the mass of the first ball thrown vertically.

For the vertical throw (Ball 1):

1. The time of flight \( T \) is given by \( T = \frac{2u}{g} \), where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.
2. The maximum height \( H_1 \) is given by \( H_1 = \frac{u^2}{2g} \).

For the angled throw (Ball 2):

1. Let \( 2m \) be the mass, with an initial speed \( v \), and angle \( \theta \).
2. The vertical component of velocity is \( v_y = v \cos \theta \).
3. Equating the times of flight \( T = \frac{2v \cos \theta}{g} \), we set both times equal, \( \frac{2u}{g} = \frac{2v \cos \theta}{g} \), leading to \( u = v \cos \theta \).
4. The maximum height \( H_2 \) is \( H_2 = \frac{(v \cos \theta)^2}{2g} = \frac{u^2}{2g} \) since \( u = v \cos \theta \).

Thus, both heights \( H_1 \) and \( H_2 \) actually come out to be equal: \( H_1 = H_2 = \frac{u^2}{2g} \). Therefore, the ratio of heights \( \frac{H_1}{H_2} \) is 1:1, meaning \( x = 1 \).

Given that the range is 1 to 1, our computed value \( x = 1 \) fits the expected range.

Answer 2

The correct answer is 1

\(∴ u1 = u2\sinθ\)
\(\frac{H_1}{H_2}=\frac{\frac{u_{1}^{2}}{2g}}{u^{2}_{2}\frac{\sin^2⁡θ}{2g}}\)
\(=\left(\frac{u_1}{u_2\sinθ}\right)^2=1\)