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Q100P

Expert-verifiedFound in: Page 39

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A**** parachutist bails out and freely falls ${\mathbf{50}}{\mathbf{}}{\mathbf{m}}$****. Then the parachute opens, and there after she decelerates at ${\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{{\mathbf{s}}}^{{\mathbf{2}}}$ ****She reaches the ground with a speed of ${\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$****. (a) How long is the parachutist in the air? (b) At what height does the fall begin?**

(a) The parachutist is in the air for $\text{17}\text{s}$ .

(b) From $\text{290}\text{m}$ the fall begins.

$\Delta y=50\text{m}$

${v}_{f}=3.0\text{m/s}$

**The problem deals with the kinematic equations of motion. Kinematics is the studies of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.**

**Formula:**

**The displacement is given by,**

**$\Delta y=\left({v}_{0}\right)\left(t\right)+\frac{1}{2}a{t}^{2}$**

**The final velocity is expressed as,**

**${{v}_{f}}^{2}={{v}_{0}}^{2}+2a\left(\Delta y\right)$**

**${v}_{f}={v}_{0}+at$**

During free fall the initial velocity parachutist is 0 m/s and acceleration of the is

$\begin{array}{rcl}a& =& g\\ & =& -9.8{\text{m/s}}^{\text{2}}\end{array}$

According to the newton’s second kinematic equation,

$\Delta y=\left({v}_{0}\right)\left(t\right)+\frac{1}{2}a{t}^{2}$

$50=0-0.5\left(9.8\right){t}^{2}$

role="math" localid="1662968239356" $t=3.19s\approx 3.2s$

The final velocity of the parachutist in free fall can be calculated as below,

According to the newton’s third kinematic equation,

${{v}_{f}}^{2}={{v}_{0}}^{2}+2a\left(\Delta y\right)$

${{v}_{f}}^{2}=2\left(9.8\right)\left(50\right)$

${v}_{f}=31.3\text{m/s}$

This final velocity for free fall of the parachutist is her initial velocity after parachute is open.

According to the newton’s first kinematic equation,

${v}_{f}={v}_{0}+at$

$3=31.3-2t$

role="math" localid="1662968456612" $t=\frac{31.3-3}{2}$

$t=14.15\phantom{\rule{0ex}{0ex}}\approx 14\text{sec}$

Therefore, the time for which the parachutist is in the air is,

$\begin{array}{rcl}3.19+14.15& =& 17.34\\ & \approx & 17\text{sec}\end{array}$

The displacement of the parachutist after parachute opened can be calculated as below,

According to the newton’s second kinematic equation,

$\Delta y=\left({v}_{0}\right)\left(t\right)+\frac{1}{2}a{t}^{2}$

$\Delta y=\left(31.3\right)\left(14\right)-\left(0.5\right)\left(2\right)\left({14}^{2}\right)$

$\begin{array}{rcl}\Delta y& =& 242\\ & \approx & 240\text{m}\end{array}$

Therefore, the height from which the fall begins is,

$\begin{array}{rcl}h& =& 50+240\\ & \approx & 290\text{m}\end{array}$

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