question_answer

Figure shows a method for measuring the acceleration due to gravity. The ball is projected upward by a "gun". The ball passes the electronic "gates" 1 and 2 as it rises and again as it falls. Each gate is connected to a separate timer. The first passage of the ball through each gate starts the corresponding timer, and the second passage through the same gate stops the timer. The time intervals \[\Delta {{t}_{1}}\] and \[\Delta {{t}_{2}}\] are thus measured. The vertical distance between the two gates is d. If \[\operatorname{d}=5m.\Delta {{t}_{1}}=3s,\Delta {{t}_{2}}=2s,\] then find the measured value of acceleration due to gravity (in\[\operatorname{m}/{{s}^{2}}\]).

A) \[8m{{s}^{-2}}\]         

B) \[4m{{s}^{-2}}\]

C) \[2m{{s}^{-2}}\]                     

D) \[1m{{s}^{-2}}\]

Correct Answer: A

Solution :

[a] \[\frac{{{v}_{1}}}{a}=\frac{\Delta {{t}_{1}}}{2}=1.5\] \[\frac{{{v}_{2}}}{a}=\frac{\Delta {{t}_{2}}}{2}=1\] \[d=\frac{v_{1}^{2}-v_{2}^{2}}{2a}\Rightarrow a=8\operatorname{m}/{{s}^{2}}\]