Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Answer to Problem 3.1P

The speed and height of the mass are as:

$\begin{array}{l}v(t)=g{t}^2\\ h(t)=\frac{1}{2}g{t}^2\end{array}$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Explanation of Solution

Given:

$m=2$,$a=g$, $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$\begin{array}{l}v=u+at\\ s=ut+\frac{1}{2}a{t}^2\end{array}$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$.

$v(t)=g{t}^2$

Similarly, for the height, that is

Since,

$s=ut+\frac{1}{2}a{t}^2$

On substituting the values

$h(t)=\frac{1}{2}g{t}^2$

The time taken to reach the platform:

$\begin{array}{l}h(t)=ut+\frac{1}{2}g{t}^2\\ \Rightarrow 6.097=0\cdot t+\frac{1}{2}\cdot 9.8\cdot t^2\\ \Rightarrow t^2=\frac{12.195}{9.81}=1\\ \Rightarrow t=1\sec \end{array}$

Similarly, The time taken to reach the ground:

$\begin{array}{l}h(t)=ut+\frac{1}{2}g{t}^2\\ \Rightarrow 9.146=0\cdot t+\frac{1}{2}\cdot 9.8\cdot t^2\\ \Rightarrow t^2=\frac{18.29}{9.81}=1.86\\ \Rightarrow t=1.36\sec \end{array}$.

Conclusion:

The speed and height of the mass are as:

$\begin{array}{l}v(t)=g{t}^2\\ h(t)=\frac{1}{2}g{t}^2\end{array}$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.