A body dropped from the top of a tower clears 7/16th of the total height of the tower in its last second of flight. The time taken by the body to reach the ground is :
According to Galileo's Ratio of Odd Number distances travelled is consecution interval of 1 sec is in the ratio of 1:3:5:7 so in 4th second distance travelled is 7/16th of total journey.
so, total time = 4 sec.
A stone dropped from the top of a tower travels \(\frac{5}{9}\) th of the height of tower during the last second of fall. Height of the tower is: (take \(g = 10\text{ m/s}^2\))
Distance in last second is \(h_{\text{last}} = \frac{5}{9}H ⇒ \text{Distance in } (t-1) \text{ seconds is } \frac{4}{9}H\). Therefore, \(\frac{\frac{1}{2}g(t-1)^2}{\frac{1}{2}gt^2} = \frac{4}{9} ⇒ \frac{t-1}{t} = \frac{2}{3} ⇒ t = 3\text{ s}\). Height \(H = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 9 = 45\text{ m}\).
Assertion (A): Two bodies of masses \(M\) and \(m\) (\(M > m\)) are allowed to fall from the same height if the air resistance force for each be the same then both the bodies will reach the earth simultaneously.
Reason (R): For same air resistance, acceleration of both the bodies will be same.
If the air resistance force \(F_{/text{air}}\) is the same for both bodies, the net force is \(mg - F_{/text{air}}\). The acceleration is \(a = g - F_{/text{air}}/m\). Since masses \(M\) and \(m\) are different, their accelerations will be different. Thus, they will not reach the earth simultaneously, and their accelerations will not be the same. Both assertion and reason are false.
Assertion (A): A body dropped from a height of \(10 \text{ m}\) from the ground will have the velocity \(5 \text{ m/s}\) at the height of \(5 \text{ m}\).
Reason (R): At the height of \(5 \text{ m}\) from the ground, the acceleration due to gravity is \(5 \text{ m/s}^2\).
Formula: \(v^2 = u^2 + 2gs\), \(g \approx 9.8 \text{ m/s}^2\).
Solution: (A) is false; for a fall of \(5 \text{ m}\) from rest, \(v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 \text{ m/s}\), not \(5 \text{ m/s}\). (R) is false; acceleration due to gravity is approximately \(9.8 \text{ m/s}^2\) or \(10 \text{ m/s}^2\), not \(5 \text{ m/s}^2\).
Assertion (A): Two balls are dropped one after the other from a tall tower. The distance between them increases linearly with time (elapsed after the second ball is dropped and before the first hits ground).
Reason (R): In given situation relative acceleration is zero, whereas relative velocity is non-zero.
Assertion (A): Let \(\Delta t\) be the time interval. The distance between them \(D(t) = \frac{1}{2}gt^2 - \frac{1}{2}g(t-\Delta t)^2 = g t \Delta t - \frac{1}{2}g (\Delta t)^2\). This is a linear function of time \(t\). So (A) is true.
Reason (R): Both balls accelerate at \(g\). Thus, their relative acceleration is \(\vec{g} - \vec{g} = \vec{0}\). The first ball has velocity \(g\Delta t\) when the second is dropped, so the relative velocity is non-zero and constant. So (R) is true.
(R) correctly explains (A): constant non-zero relative velocity results in linear increase in relative distance.
A ball is thrown vertically downward with a velocity of \(20\text{ m/s}\) from the top of a tower. It hits the ground after some time with a velocity of \(80\text{ m/s}\). The height of the tower is : \((g = 10\text{ m/s}^2)\)
(2020)
Concept: Equations of motion under gravity.
Formula: \(v^2 = u^2 + 2gh\).
Solution: Given \(u = 20\text{ m/s}\,\text{ }v = 80\text{ m/s}\,\text{ }g = 10\text{ m/s}^2\). Substituting these values: \(80^2 = 20^2 + 2(10)h\). \(6400 = 400 + 20h\). \(6000 = 20h\) => \(h = 300\text{ m}\).
A person sitting in the ground floor of a building notices through the window, of height \(1.5\text{ m}\), a ball dropped from the roof of the building crosses the window in \(0.1\text{ s}\). What is the velocity of the ball when it is at the topmost point of the window? \((g = 10\text{ m/s}^2)\)
(2020-Covid)
Concept: Equations of motion under gravity for a specific interval.
Formula: \(h = ut + \frac{1}{2}gt^2\).
Solution: Let \(u\) be velocity at window top. Given \(h=1.5\text{ m}\,\text{ }t=0.1\text{ s}\,\text{ }g=10\text{ m/s}^2\). \(1.5 = u(0.1) + \frac{1}{2}(10)(0.1)^2\). \(1.5 = 0.1u + 0.05\). \(1.45 = 0.1u\) => \(u = 14.5\text{ m/s}\).
A stone falls freely under gravity. It covers distances \(h_1, h_2\) and \(h_3\) in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between \(h_1, h_2\) and \(h_3\) is:
(2013)
Concept: Distances covered by a freely falling body in equal successive time intervals.
Rule: For a body falling from rest, distances in successive equal time intervals are in ratio 1:3:5:...
Solution: \(h_1:h_2:h_3 = 1:3:5\). This implies \(h_2 = 3h_1\) and \(h_3 = 5h_1\). Therefore, \(h_1 = h_2/3 = h_3/5\).
A boy standing at the top of a tower of \(20\text{ m}\text{ height drops a stone. Assuming } g = 10\text{ m/s}^2\text{, the velocity with which it hits the ground is:}\)
[2011 Pre]
Concept: Free fall under gravity.
Formula: \(v^2 = u^2 + 2gh\).
Solution: Given \(u=0\) (dropped), \(h=20\text{ m}\,\text{ }g=10\text{ m/s}^2\). \(v^2 = 0^2 + 2(10)(20) = 400\). So, \(v = \sqrt{400} = 20\text{ m/s}\).
Two bodies, A (of mass \(1\text{ kg}\)) and B (of mass \(3\text{ kg}\)) are dropped from heights of \(16\text{ m}\) and \(25\text{ m}\), respectively. The ratio of the time taken by them to reach the ground is:
[2006]
Concept: Free fall under gravity.
Formula: Distance \(h = \frac{1}{2}gt^2\) ⇒ time \(t = \sqrt{\frac{2h}{g}}\), so \(t \propto \sqrt{h}\)
For body A, \(h_A = 16\text{ m}\); for body B, \(h_B = 25\text{ m}\).
Ratio: \(t_A/t_B = \sqrt{h_A/h_B} = \sqrt{16/25} = 4/5\).