Two bodies of different masses \(m_a\) and \(m_b\) are dropped from two different heights, viz, an and b. The ratio of time taken by the two to drop through these distances is :
1. a : b
2. \( \frac{m_a}{m_b} : \frac{b}{a} \)
3. \( \sqrt{a} : \sqrt{b} \)
4. \(a^2 : b^2 \)
View Answer
The time taken to fall through height \(h\) from rest under gravity is \(t = \sqrt{\frac{2h}{g}}\). Since \(t \propto \sqrt{h}\), the ratio of times for heights \(a\) and \(b\) is \(\sqrt{a} : \sqrt{b}\).
If a body A of mass M is thrown with velocity \(v\) at an angle \(30^\circ\) to the horizontal and another body B of same mass is thrown with same speed at an angle of \(60^\circ\) to the horizontal, the ratio of range of A and B will be :
1. \( 1 : \sqrt{3} \)
2. \( \sqrt{3} : 1\)
4. 1 : 1
View Answer
Horizontal range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For complementary projection angles \(\theta\) and \(90^\circ - \theta\) (like \(30^\circ\) and \(60^\circ\)), the horizontal ranges are equal. Thus, the ratio of range is \(1 : 1\).