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A boy sees a ball go up and then down through a window 2.45m high if the total time the ball is in sight is 1s. The height down the window the ball raises is approximately.
A) 2.45m
B) 4.9m
C) 0.3m
D) 0.49m

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Last updated date: 27th Jul 2024
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Answer
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Hint: The solution to this problem is obtained by utilizing the second equation of motion and also by the speed and direction relationship. Equations of motion are the equations which describe the behavior of a physical system in terms of its motion as a function of time. Three equations of motion derive displacement, velocity and time.

Complete step-by-step solution:
 Equation of motion is given as:
$s=ut-\dfrac{1}{2}g{{t}^{2}}$ $\cdots \cdots (1)$
Where $u$ = speed when the ball passes the bottom of the window.
              t= $\dfrac{1}{2}\sec $ =Time duration for travelling window height one way.
              g=$9.8m{{s}^{-2}}$
Substitute these values in equation (1)
$\begin{align}
  & 2.45=u\times \dfrac{1}{2}-\dfrac{1}{2}\times 9.8\times {{\dfrac{1}{2}}^{2}} \\
 & u=7.35m{{s}^{-1}} \\
\end{align}$
The total height at which the ball rises from the bottom of the window is :
$\begin{align}
  & {{s}_{b}}=\dfrac{{{7.35}^{2}}}{2\times 9.8} \\
 & {{s}_{b}}=2.75625m \\
\end{align}$
Ball rises $= {{X}_{f}}-{{X}_{i}}=2.75625-2.45 =0.30625$m
So, The height the ball rises from the bottom of the window to the top of the window =0.3m
The correct option is C
Additional information: Displacement is defined as the process in which objects' positions are changed and in displacement the initial position of objects are changed. Displacement is also defined as change in initial position of objects to the final position and displacement is denoted as S.
Displacement $S={{X}_{f}}-{{X}_{i}}$
Where \[{{X}_{f}}\] =final position
              ${{X}_{i}}$ =initial position
Velocity is defined as the rate of change of displacement with respect to time and in kinematics velocity is a fundamental concept.SI unit of velocity is $m{{s}^{-1}}$ velocity tracking is the measure of velocity.
Velocity (v) =$\dfrac{\Delta S}{\Delta t}$

Note: Students, the displacement is a vector quantity which has both magnitude and direction and displacement is measured in terms of meters. The dimensional formula of displacement is ${{M}^{0}}{{L}^{1}}{{T}^{0}}$ and displacement plays a very important role while determining velocity (v). Velocity is also a vector quantity.