Question

A bolt is dropped from a bridge under construction, falling 90 mto the valley below the bridge. (a) In how much time does it pass through the last 20%of its fall? What is its speed? (b) When it begins the last20% of its fall. (c) When it reaches the valley beneath the bridge?

Short answer

(a)Time required to pass through the last 20% of the fall is 0.45 s .

(b)Velocity when it starts the last 20% of the fall is 38 m/s .

(c)Final velocity when it is just above the ground is 42 m/s .

Step-by-step solution

Understanding the concept

The height of the bridge above the ground is given.The kinematic equations give the relationship between initial, final velocity, acceleration, displacement, and time.The required velocities can be determined by using kinematics equations.

The Initial velocity is $v_0=0\mathrm{m}/\mathrm{s}$.

The acceleration due to gravity is$g=9.8\mathrm{m}/{\mathrm{s}}^2$.

The distancetraveled by the bolt is, $s=90\mathrm{m}$.

The kinematic equation for the final velocity in terms of initial velocity, acceleration, and time is,

$v=v_0+at$ (i)

The kinematic equation for the distance traveled is,

$s=v_0+\frac{1}{2}a{t}^2$ (ii)

(a) Determination of time required to pass through the last of the fall

Calculation of the time required to cover the total distance$s_1$

$$\begin{gathered} s_1=v_0{t}_1+\frac{1}{2}a{t_1}^2 \\ 90m=0+\frac{1}{2}\times 9.8m/s^2\times {t_1}^2 \\ {t_1}^2=\frac{90\times 2}{9.8} \\ {t}_1=\sqrt{\frac{180}{90}} \\ =4.28\mathrm{s} \end{gathered}$$

Therefore, the time required to cover the total distance is $4.28\mathrm{s}$.

Now, to calculate the time required to cover first 80% of the distance, find the distance that needs to be travelled.

$$\begin{gathered} s_2=\frac{80}{100}\times 90\mathrm{m} \\ =72\mathrm{m} \end{gathered}$$

Initial Velocity $v_0=0\frac{\mathrm{m}}{\mathrm{s}}$

Acceleration is gravitational acceleration that is equal to$9.8\mathrm{m}/{\mathrm{s}}^2$.

$$\begin{gathered} s_2=v_{0}t+\frac{1}{2}a{t_2}^2 \\ 72m=0+\frac{1}{2}\times 9.8m/s^2\times {t_2}^2 \\ \frac{72\times 2}{9.8}={t_2}^2 \\ {t}_2=\sqrt{\frac{144}{9.8}} \\ =3.83\mathrm{s} \end{gathered}$$

Therefore, time required to cover 20% of the distance is equal to the difference between the time required to cover 100% of distance and the time required to cover 80% of the distance.

$$\begin{gathered} ∆t=t_1-t_2 \\ =4.28\mathrm{s}-3.83\mathrm{s} \\ =0.45\mathrm{s} \end{gathered}$$

Therefore, time required to pass through the last 20% of the fall is$0.45\mathrm{s}$

(b) Determination of velocity when last 20% of the fall started

The time $t_2$when last 20% of the fall starts is3.83 s

Putting the values in equation (i),

$$\begin{gathered} v_2=v_0+a{t}_2 \\ =0+9.8\mathrm{m}/{\mathrm{s}}^2\times 3.83\mathrm{s} \\ =37.5\mathrm{m}/\mathrm{s}\approx 38\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, velocity when it starts the last 20% of the fall is 38 m/s .

(c) Determination of final velocity

The time when the bolt reaches valley beneath the bridge is$t_1=4.28\mathrm{s}$ . Substitute the value in equation (i) to find the final velocity.

$$\begin{gathered} v_1=v_0+a{t}_1 \\ =0+9.8\mathrm{m}/{\mathrm{s}}^2\times 4.28\mathrm{s} \\ =41.9\approx 42\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, final velocity when it is just above the ground is $42\mathrm{m}/\mathrm{s}$.