Chapter 6, Problem 47A

a.

To determine

Time taken bythe car to fall.

Answer to Problem 47A

Time taken by the car to fall is, $t=0.5\text{ s}$ .

Explanation of Solution

Given :

Height of the table, $h=1.225\text{ m}$

Car lands 0.4 m from the base of table, $d=0.4\text{ m}$

Formula used :

Second equation of motion (considering the effect of gravity)

  $h=ut+\frac{1}{2}g{t}^2$

Where,

u is initial velocity

g is acceleration due to gravity

t is time.

Calculation:

When car leaves the table initial velocity is, $u=0\text{ m/sec}$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Now, substitute the value of u, g and hin equation of motion.

  $\begin{array}{l}1.225=0\times t+\frac{1}{2}\times 9.8\times t^2\\ 4.9t^2=1.225\\ t^2=0.25\\ t=0.5\text{ s}\end{array}$

Conclusion:

Therefore, time taken to fall the car is 0.5 s.

b.

To determine

Velocity of car on table.

Answer to Problem 47A

Velocity of car on table is, $v=8\text{ m/s}$ .

Explanation of Solution

Given :

Height of the table, $h=1.225\text{ m}$

Car lands 0.4 m from the base of table, $d=0.4\text{ m}$

Formula used :

Velocity is defined as,

  $v=\frac{d}{t}$

Where,

d is distance

t is time.

Calculation:

Substitute the value of d and t

  $\begin{array}{c}v=\frac{4}{0.5}\\ =8\text{ m/s}\end{array}$

Conclusion:

Therefore, velocity of car on the table is 8 m/s.