(B) For a freely falling body, the equation of motion for velocity is given by $v = u + gt$. Since the initial speed $u = 0$, the equation becomes $v

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a straight line passing through the origin.

(B) For a freely falling body, the equation of motion for velocity is given by $v = u + gt$. Since the initial speed $u = 0$, the equation becomes $v = gt$. Here, $g$ is the acceleration due to gravity, which is a constant ($g \approx 9.8 \ m/s^2$ or $10 \ m/s^2$). This equation $v = gt$ is of the form $y = mx$, where $y$ represents speed, $x$ represents time, and $m = g$ is the slope. Since there is no constant term (intercept), the graph is a straight line passing through the origin $(0, 0)$. Therefore, the correct option is $B$.

Class 11 Physics Motion in Straight Line Motion Under Gravity

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The nature of a graph drawn for a freely falling body with time on the $x$-axis and speed on the $y$-axis is (Assuming initial speed to be zero.)

TS EAMCET 2019 All TS EAMCET

A
a straight line with positive $y$-axis intercept.
B
a straight line passing through the origin.
C
a parabola.
D
a straight line parallel to $y$-axis with positive $x$-axis intercept.

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Chapter

Motion in Straight Line

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Motion Under Gravity

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$A$ balloon rises up with a constant net acceleration of $10\,m/s^2$. After $2\,s$,a particle drops from the balloon. After a further $2\,s$,match the following columns. (Take $g=10\,m/s^2$)

Column $I$ Column $II$

$(A)$ Height of particle from ground $(p)$ $0$

$(B)$ Speed of particle $(q)$ $10$ $SI$ unit

$(C)$ Displacement of particle $(r)$ $40$ $SI$ unit

$(D)$ Acceleration of particle $(s)$ $20$ $SI$ unit

Column $I$	Column $II$
$(A)$ Height of particle from ground	$(p)$ $0$
$(B)$ Speed of particle	$(q)$ $10$ $SI$ unit
$(C)$ Displacement of particle	$(r)$ $40$ $SI$ unit
$(D)$ Acceleration of particle	$(s)$ $20$ $SI$ unit

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The nature of a graph drawn for a freely falling body with time on the $x$-axis and speed on the $y$-axis is (Assuming initial speed to be zero.)

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