A ball is thrown vertically upward. It has a speed of \(10\text{ m/s}\) when it has reached one half of its maximum height. How high does the ball rise? (Taking \(‘g’ = 10\text{ m/s}^2\)\)
(2005)
1. \(6\text{ m}\)
2. \(10\text{ m}\)
3. \(14\text{ m}\)
4. \(18/text{ m}\)
View Answer
Concept: Vertical motion under gravity. Let \(H\) be maximum height, \(u\) be initial velocity.
Formula: \(v^2 = u^2 - 2gh\). At \(H\), \(v=0 \Rightarrow u^2 = 2gH\).
At \(H/2\), \(10^2 = u^2 - 2g(H/2) = u^2 - gH\).
Substitute \(u^2 = 2gH\): \(100 = 2gH - gH = gH\).
Given \(g=10\text{ m/s}^2\), so \(100 = 10H \Rightarrow H = 10\text{ m}\).
A man throws ball with the same speed vertically upwards one after the other at an interval of \(2\text{ seconds}\)). What should be the speed of the throw so that more than two balls are in the sky at any time? (Given \(g = 9.8 m/s^2\)
(2003)
1. More than \(19.6\text{ m/s}\)
2. At least \(9.8\text{ m/s}\)
3. Any speed less than \(19.6\text{ m/s}\)
4. Only with speed \(19.6\text{ m/s}\)
View Answer
Concept: Time of flight for vertical motion. Let \(\Delta t = 2\text{ s}\)) be the throwing interval.
Formula: Time of flight \(T = 2u/g\).
For more than two balls to be in the air, the time of flight of each ball must be greater than twice the interval: \(T > 2\Delta t\).
So, \(2u/g > 2 \times 2 = 4\text{ s}\)).
\(u > 2g = 2 \times 9.8 = 19.6\text{ m/s}\).
A body dropped from a height \(h\) with initial velocity zero, strikes the ground with a velocity \(3\text{ m/s}\)). Another body of same mass dropped from the same height \(h\) with an initial velocity of \(4\text{ m/s}\)). The final velocity of second mass, with which it strikes the ground is:
(1996)
1. \(5\text{ m/s}\)
2. \(12\text{ m/s}\)
3. \(3\text{ m/s}\)
4. \(4\text{ m/s}\)
View Answer
Concept: Equations of motion under constant gravity.
Formula: \(v_f^2 = v_i^2 + 2gh\).
For the first body: \(v_i = 0\), \(v_f = 3\text{ m/s}\). So, \(3^2 = 0^2 + 2gh \Rightarrow 2gh = 9\).
For the second body: \(v_i = 4\text{ m/s}\). The final velocity is \(v_f'\).
\((v_f')^2 = 4^2 + 2gh = 16 + 9 = 25\).
Therefore, \(v_f' = \sqrt{25} = 5\text{ m/s}\).