91影视

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)

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

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

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.

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

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

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

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

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.

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

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

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

Putting the values in equation (i),

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

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

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.

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

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