Question

The window of the fourth floor of SANKALP building is 5 m high. A man looking out of the window sees an object moving up and down the height of window for 2 sec. Find the height that the object reaches from the top end of the window.

Answer 1

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Answer 2

Let the height of the window be $H = 5$ m. The object is seen moving up and down the height of the window for a total time of $T = 2$ s. Let $t_{up}$ be the time taken for the object to move upwards through the window, and $t_{down}$ be the time taken for the object to move downwards through the window. The total time spent within the window is $t_{up} + t_{down} = 2$ s. Due to the symmetry of motion under gravity, the time taken to go up through the window is equal to the time taken to come down through the window, provided the velocities at the top and bottom are the same in magnitude for both upward and downward journeys. So, $t_{up} = t_{down} = T/2 = 2 \text{ s} / 2 = 1$ s.

Let $v_B$ be the velocity of the object at the bottom of the window (when moving upwards) and $v_T$ be the velocity of the object at the top of the window (when moving upwards). Consider the upward motion through the window. The displacement is $H = 5$ m, the time taken is $t_{up} = 1$ s, the initial velocity is $v_B$, and the final velocity is $v_T$. The acceleration is $-g$ (taking upward direction as positive). Using the equation of motion $s = ut + \frac{1}{2}at^2$: $H = v_B t_{up} + \frac{1}{2}(-g)t_{up}^2$ $5 = v_B (1) - \frac{1}{2}g(1)^2$ $5 = v_B - \frac{g}{2}$ $v_B = 5 + \frac{g}{2}$

Using the equation of motion $v = u + at$: $v_T = v_B + (-g)t_{up}$ $v_T = v_B - g(1)$ $v_T = v_B - g$ Substitute the expression for $v_B$: $v_T = (5 + \frac{g}{2}) - g = 5 - \frac{g}{2}$

The object reaches its maximum height when its velocity becomes 0. Let $h$ be the height the object reaches above the top end of the window. The object leaves the top of the window with an upward velocity $v_T$. It travels upwards until its velocity becomes 0. Using the equation of motion $v^2 = u^2 + 2as$ for the motion from the top of the window to the maximum height: Initial velocity $u = v_T$ (upwards) Final velocity $v = 0$ (at maximum height) Displacement $s = h$ (upwards) Acceleration $a = -g$ $0^2 = v_T^2 + 2(-g)h$ $0 = v_T^2 - 2gh$ $2gh = v_T^2$ $h = \frac{v_T^2}{2g}$

Substitute the expression for $v_T$: $h = \frac{(5 - g/2)^2}{2g}$

The problem does not specify the value of $g$. In the absence of a specified value, we can assume $g = 10 \text{ m/s}^2$ as is common in many problems. If $g = 10 \text{ m/s}^2$: $v_T = 5 - \frac{10}{2} = 5 - 5 = 0 \text{ m/s}$. If $v_T = 0$, the object's velocity is 0 at the top of the window. This means the object reaches its maximum height exactly at the top of the window. The height reached from the top end of the window is $h = \frac{0^2}{2g} = 0$ m.