Chapter 6, Problem 45A

To determine

The far from the cliff key looks.

Answer to Problem 45A

Key will look at a distance of 28.88 m from the base of cliff.

Explanation of Solution

Given:

Horizontal velocity of car key $v_x=8\text{ m/s}$

Height from which key has thrown $h=64\text{ m}$

Formula Used:

The third equation of motion:

  $s=ut+\frac{1}{2}a{t}^2$

Where, u is the initial velocity, t is the time, a is the acceleration and s is the displacement.

Calculations:

When the key is thrown, its initial velocity is $u=0\text{ m/s}$ .

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

Now, substitute the values and solve:

  $\begin{array}{l}64=0\times t+\frac{1}{2}\times 9.8\times t^2\\ 4.9t^2=64\\ t^2=13.06\\ t=3.61\text{ s}\end{array}$

There is no acceleration in horizontal direction. So, the horizontal distance is:

  $x=v_{x}t$

Substitute the values and solve:

  $\begin{array}{c}x=8\times 3.61\\ =28.88\end{array}$

Conclusion:

Therefore, key will look at a distance of 28.88 m.