Question

Extreme-sports enthusiasts have been known to jump off the top of El Captain, a sheer granite cliff of height 910 m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Captain with speed$4.0\mathrm{m}/\mathrm{s}$and enjoys a free fall until she is 150 m above the valley floor, at which time she opens her parachute (Fig. 3-37). (a) How long is the jumper in free fall? Ignore air-resistance. (b) It is important to be as far away from the cliff as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute?

Short answer

(a) The time taken by the jumper is 12.45 s. (b) The horizontal distance between the cliff and the jumper before opening the parachute is 49.8 m.

Step-by-step solution

Step 1. Meaning of free fall motion

The motion of an object that is exclusively under the action of gravity is known as free fall motion. This object falls downward, experiencing acceleration due to gravity. Some examples are a spacecraft moving in an orbit, meteor shower toward the Earth, sky diving, etc.

Step 2. Identification of the given data

The horizontal speed of the jumper is $u_{\mathrm{x}}=4.0\mathrm{m}/\mathrm{s}$.

The height of the cliff from the valley floor is $H=910\mathrm{m}$.

The height at which the jumper opens the parachute is $h=150\mathrm{m}$.

Let the valley floor or ground level be at $y=0$, and the positive y-direction be upward.

The initial vertical velocity is $u_{\mathrm{y}}=0$.

The vertical acceleration experienced by the jumper is $a_{\mathrm{y}}=-g=-9.8\mathrm{m}/{\mathrm{s}}^2$.

Step 3. (a) Determination of the time taken during free fall

The kinematic equation of motion along the y-direction is given as:

$\begin{array}{c}y=u_{\mathrm{y}}t+\frac{1}{2}{a}_{\mathrm{y}}{t}^2\\ h-H=0+\frac{1}{2}\left(-g\right)t^2\end{array}$

Substituting the given numerical values in the above expression for the time taken by the jumper to remain in free fall, you get:

$\begin{array}{c}\left(150-910\right)=0+\frac{1}{2}\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t=\sqrt{\frac{760\mathrm{m}}{4.9\mathrm{m}/{\mathrm{s}}^2}}\\ =12.45\mathrm{s}\end{array}$

Thus, the time taken by the jumper in the free fall is 12.45 s.

Step 4. (b) Determination of the horizontal distance between the cliff and the jumper before opening the parachute

The horizontal motion of the jumper at a constant horizontal speed is given by:

$\begin{array}{c}x=v_{\mathrm{x}}t\\ =\left(4.0\mathrm{m}/\mathrm{s}\right)\left(12.45\mathrm{s}\right)\\ =49.8\mathrm{m}\end{array}$