Question 112: Motion in 1D

Question

A body falling under gravity covers two points $$A$$ and $$B$$ separated by $$80 \mathrm{~m}$$ in $$2 \mathrm{~s}$$. The distance of upper point A from the starting point is _________ $$\mathrm{m}$$ (use $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$).

Accepted Answer

To find the distance of the upper point A from the starting point, we need to first understand the motion of a freely falling body under the influence of gravity. The body falling under gravity is an example of uniformly accelerated motion with the acceleration equal to the acceleration due to gravity, which is given as $ g = 10 	extrm{m/s}^2 $.
We will consider point A where the body was at time $ t_1 $ and point B where the body was at time $ t_2 = t_1 + 2 	extrm{s} $. The displacement in these 2 seconds is given to be 80 meters.
For an object under constant acceleration, the displacement $ s $ can be found using the equation:
$ s = ut + \frac{1}{2}at^2 $
where
$ u $ is the initial velocity,
$ t $ is the time,
$ a $ is the acceleration,
$ s $ is the displacement.
Since the body falls under gravity, its initial velocity at the starting point is 0 ($ u = 0 $), thus the equation simplifies to:
$ s = \frac{1}{2}gt^2 $
Let's denote the distance of point A from the starting point as $ s_A $ and the distance of point B as $ s_B $. We know the body covers $ 80 $ meters in $ 2 $ seconds from A to B, so we can write:
$ s_B - s_A = 80 	extrm{m} $
We can calculate the distance covered till point B (in time $ t_2 = t_1 + 2 $) as:
$ s_B = \frac{1}{2}g(t_1 + 2)^2 $
And the distance covered till point A (in time $ t_1 $) as:
$ s_A = \frac{1}{2}gt_1^2 $
Now, substituting $ s_B $ and $ s_A $ in our difference equation:
$ \frac{1}{2}g(t_1 + 2)^2 - \frac{1}{2}gt_1^2 = 80 $
$ g \left( \frac{1}{2}(t_1 + 2)^2 - \frac{1}{2}t_1^2 ight) = 80 $
$ 5((t_1 + 2)^2 - t_1^2) = 80 $
$ (t_1 + 2)^2 - t_1^2 = 16 $
$ t_1^2 + 4t_1 + 4 - t_1^2 = 16 $
$ 4t_1 + 4 = 16 $
$ 4t_1 = 16 - 4 $
$ 4t_1 = 12 $
$ t_1 = 3 	extrm{s} $
Therefore, time $ t_1 $ at point A is $ 3 $ seconds. To find $ s_A $, we can use the above simplified motion equation:
$ s_A = \frac{1}{2}gt_1^2 $
$ s_A = \frac{1}{2} 	imes 10 	imes 3^2 $
$ s_A = 5 	imes 9 $
$ s_A = 45 	extrm{m} $
So, the distance of the upper point A from the starting point is $ \mathbf{45 	extrm{m}} $.
Alternate Method :
From $\mathrm{A} ightarrow \mathrm{B}$
$$
\begin{aligned}
& -80=-\mathrm{v}_1 \mathrm{t}-\frac{1}{2} 	imes 10 \mathrm{t}^2 \\
& -80=-2 \mathrm{v}_1-\frac{1}{2} 	imes 10 	imes 2^2 \\
& -80=-2 \mathrm{v}_1-20 \\
& -60=-2 \mathrm{v}_1 \\
& \mathrm{v}_1=30 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
From $\mathrm{O}$ to $\mathrm{A}$
$$
\mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gS}
$$
$$ \Rightarrow $$ $$
30^2=0+2 	imes(-10)(-\mathrm{S})
$$
$$ \Rightarrow $$ $$
900=20 \mathrm{~S}
$$
$$ \Rightarrow $$ $$
\mathrm{S}=45 \mathrm{~m}
$$

Motion in 1D

A body falling under gravity covers two points $A and B separated by 80 \mathrm{~m} in 2 \mathrm{~s}. The distance of upper point A from the starting point is _________ \mathrm{m} (use \mathrm{g}=10 \mathrm{~ms}^{-2}$).

Integer / Numeric Answer Type