Chapter 6.1, Problem 1PP

a.

To determine

Time taken by the stone to reach the bottom of the cliff.

Answer to Problem 1PP

Time taken by stone to reach bottom of cliff, $t=4\text{ s}$ .

Explanation of Solution

Given :

Horizontal velocity of the stone, $v_x=5\text{ m/s}$

Height of the cliff, $h=78.4\text{ m}$

Formula used :

Second equation of motion.

  $h=ut+\frac{1}{2}g{t}^2$

Here, u is initial velocity, g is acceleration due to gravity, and t is time.

Calculation:

When stone is thrown its initial velocity, $u=0\text{ m/sec}$

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

Now, substitute the value of u, g and h and solve.

  $\begin{array}{l}78.4=0\times t+\frac{1}{2}\times 9.8\times t^2\\ 4.9t^2=78.4\\ t^2=16\\ t=4\text{ s}\end{array}$

Conclusion :

Therefore, time taken by stone to reach the cliff is 4 s.

b.

To determine

The distanceof stone hitting the ground from the base of the cliff.

Answer to Problem 1PP

Stone will hit the ground at distance of $d=20\text{ m}$ .

Explanation of Solution

Given :

Horizontal velocity of the stone, $v_x=5\text{ m/s}$

Height of the cliff, $h=78.4\text{ m}$

Formula used :

  $d=vt$

Here, d is distance, v is speed and t is time.

Calculation:

From part a, $t=4\text{ s}$

Now, substitute the value of v and t .

  $\begin{array}{c}d=5\times 4\\ =20\text{ m}\end{array}$

Conclusion :

Therefore, stone will hit the ground at a distance of 20 m from the base of cliff.

c.

To determine

Horizontal and vertical component of velocity stone just before it hits the ground.

Answer to Problem 1PP

Horizontal component of velocity is $v_x=5\text{ m/s}$ and vertical component of velocity is $v_y=39.2\text{ m/s}$ .

Explanation of Solution

Given :

Horizontal velocity of the stone, $v_x=5\text{ m/s}$

Height of the cliff, $h=78.4\text{ m}$

Formula used :

  $v=u+gt$

Here, v is final velocity, u is initial velocity and g is acceleration due to gravity.

Calculation:

As there is no influence of gravity on horizontal therefore horizontal velocity will remain same as $v_x=5\text{ m/s}$ when stone hits the ground.

Initial velocity in vertical direction, $u_{iy}=0\text{ m/s}$

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

Now, substitute the value of $u_{iy}$ , g and t.

  $\begin{array}{c}{v}_y=0+9.8\times 4\\ =39.2\text{ m/s}\end{array}$

Conclusion:

Therefore, horizontal component of velocity when stone hits the ground is 5 m/s and vertical component of velocity when stone hits the ground is 39.2 m/s.