Question

Question: At time t=0, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-33 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. The scaling is set by t_s=2s. With approximately what speed is apple 2 thrown down?

Short answer

The speed of apple 2 when it is thrown down is ${\mathrm{V}}_2=9.6\mathrm{m}/\mathrm{s}$

Step-by-step solution

Given information

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)

Understanding the concept of free fall

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,

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

Determination of Height (Δy) from which the apple was dropped

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

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

Determination of speed of apple 2 when it is thrown down

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

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

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

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.