Chapter 2, Problem 34QAP

To determine

The values of the given table.

Answer to Problem 34QAP

t(s)	y(m)	v(m/s)	a(m/s2)
0	0	0	9.80
1	4.90	9.80	9.80
2	19.6	19.6	9.80
3	44.1	29.4	9.80
4	78.4	39.2	9.80
5	122	49.0	9.80
10	490	98.0	9.80

Explanation of Solution

Given info:

  $a=9.8{\text{ m/s}}^2$

  $t=0,1,2,3,4,5,10\text{ s}$

Formula used:

  $y=v_{0}t+\frac{1}{2}a{t}^2$

  $v=v_0+at$

Calculation:

Assuming that the direction of the displacement of the coconut and the gravitational acceleration are same,

  $\begin{array}{l}{y}_{(t=1)}=0\times 1+\frac{1}{2}\times 9.80\times 1^2=4.90\text{ m}\\ y_{(t=2)}=0\times 2+\frac{1}{2}\times 9.80\times 2^2=19.6\text{ m}\\ y_{(t=3)}=0\times 3+\frac{1}{2}\times 9.80\times 3^2=44.1\text{ m}\\ y_{(t=4)}=0\times 4+\frac{1}{2}\times 9.80\times 4^2=78.4\text{ m}\\ y_{(t=5)}=0\times 5+\frac{1}{2}\times 9.80\times 5^2=122\text{ m}\\ y_{(t=10)}=0\times 10+\frac{1}{2}\times 9.80\times {10}^2=490\text{ m}\end{array}$

  $\begin{array}{l}{v}_{(t=1)}=0+9.80\times 1=9.80\text{ m/s}\\ v_{(t=2)}=0+9.80\times 2=19.6\text{ m/s}\\ v_{(t=3)}=0+9.80\times 3=29.4\text{ m/s}\\ v_{(t=4)}=0+9.80\times 4=39.2\text{ m/s}\\ v_{(t=5)}=0+9.80\times 5=49.0\text{ m/s}\\ v_{(t=10)}=0+9.80\times 10=98.0\text{ m/s}\end{array}$

Conclusion:

t(s)	y(m)	v(m/s)	a(m/s2)
0	0	0	9.80
1	4.90	9.80	9.80
2	19.6	19.6	9.80
3	44.1	29.4	9.80
4	78.4	39.2	9.80
5	122	49.0	9.80
10	490	98.0	9.80