Chapter 8.8, Problem 26E

a.

To determine

To find: the distance between the stones after landing.

Answer to Problem 26E

106.75 feet

Explanation of Solution

Given information:

Velocity of stones thrown horizontally = 35 ft/s

Height of cliff from the ground = 150 feet

Calculation:

Let assume one stone $\left(A\right)$ will drop straight down and stop immediately where it lands; and the other stone $\left(B\right)$ be thrown at exact horizontal direction. So the stone $B$ will be expected to fall 150 feet at an initial vertical velocity and vertical acceleration of zero and 32.2 ${\text{ft/s}}^2$ .

The time taken to land will be evaluated as follows:

  $\begin{array}{c}y=y_o-\frac{1}{2}g{t}^2\\ 0=150-\frac{1}{2}\left(32.2\right)t^2\\ -150=-16.1t^2\\ t^2=9.3\\ t\approx 3.05\text{ s}\end{array}$

Now evaluate the distance travelled by stone at a constant vertical velocity of $35\text{ ft/s}$ as shown below.

  $\begin{array}{c}x=x_o+vt\\ =0+\left(35\right)\left(3.05\right)\\ =106.75\text{ ft}\end{array}$

Thus the distance between the stones after landing is 106.75 feet.

b.

To determine

To state: whether the stones will land at same time. Also explain the reason.

Answer to Problem 26E

Yes

Explanation of Solution

Given information:

Velocity of stones thrown horizontally = 35 ft/s

Height of cliff from the ground = 150 feet

Time of landing = 3.05 seconds

Calculation:

If the air resistance is ignored, then the rate of vertical acceleration for two bodies is exactly same. The time of landing of each stone evaluated in part a. is 3.05 seconds. Thus the stones will land at same time