91Ó°ÊÓ

Acceleration (a)=Gravitational Acceleration= 9.8 m/s²

Time taken by apple 1 to reach ground (t)=2sec

Initial Speed of apple 1 (V₁)=0 m/sec

Time taken by apple 2 to reach the ground=2.25s (from the graph)

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Theseequations are used to find the height from which the first apple was dropped.Since it is known the time taken by the apple to reach the ground, free fall acceleration and initial velocity of the apple, and using the information about the time on the graph, find the time it took for the second apple to reach the ground. From this information find the initial velocity of the second apple.

Formula:

The displacement is given by,

$∆\mathrm{y}={\mathrm{V}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

Consider the direction of acceleration downward as positive and upward as negative.

As the apple is falling down, the acceleration of the apple would be gravitational acceleration.

So, using the kinematic equation

$$\begin{gathered} ∆\mathrm{y}=\left({\mathrm{V}}_1×\mathrm{t}\right)+\frac{1}{2}×\mathrm{a}×{\mathrm{t}}^2 \\ ∆\mathrm{y}=0×2\mathrm{s}+\frac{1}{2}×9.8\mathrm{m}/{\mathrm{s}}^2×2^2{\mathrm{s}}^2 \\ ∆\mathrm{y}=19.6\mathrm{m} \end{gathered}$$

From the graph the apple 2 is thrown 1 sec after apple 1 is thrown. So, the time taken by apple 2 to reach the ground is

From the kinematic equation

$$\begin{gathered} ∆\mathrm{y}={\mathrm{V}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ 19.6m=\left(V_2×1.25s\right)+\frac{1}{2}×9.8m/s^2×{\left(1.25s\right)}^2 \end{gathered}$$

${\mathrm{v}}_2=\left(\frac{19.6\text{m}-7.66\text{m}}{1.25\text{s}}\right)$

V₂=9.552 m/s

=9.6 m/s

So, the velocity of the second apple would be 9.6 m/s in the downward direction.