Free Fall!

Classical Mechanics Level 2

At time $t=0$ , apple 1 is dropped from a bridge onto a roadway beneath the bridge. Somewhat later, apple 2 is thrown down from the same height. The figure gives the vertical positions y of the apples versus t during falling, until both apples have hit the roadway. With approximately what speed is apple 2 thrown down? Take $g=9.8 \text{ m/s}^2$ .

7.8 m/s 9.6 m/s 5.2 m/s 11.2 m/s

2 solutions

Muhammad Erfan
Jul 20, 2015

the distance of the drop for the first apple ( h ):

$h={ v }_{ 01 }t+\frac { 1 }{ 2 } g{ t }^{ 2 }$ since the first apple has dropped, so ${ v }_{ 0 }=0$ $h=\frac { 1 }{ 2 } g{ t }^{ 2 }=\frac { 1 }{ 2 } \left( 9.8 \right) \left( { 2 }^{ 2 } \right) =19.6\quad m$

the time of the second apple is 1.25 s

the speed of the second apple:

${ v }_{ 02 }=\frac { \left( 19.6 \right) -\left( \frac { 1 }{ 2 } \right) \left( 9.8 \right) { \left( 1.25 \right) }^{ 2 } }{ \left( 1.25 \right) } =9.555\quad \approx \quad 9.6\quad \frac{m}{s}$

so the answer is 9.6 m/s

Anindya Mahajan
Jul 15, 2015

using the 2nd eqn. of motion for the first apple, the distance of the drop is calculated at 19.6m. now, using the 2nd eqn. of motion for the second apple, one can easily arrive at the initial velocity with which it was thrown 'u'=9.555 m/s. the closest answer in the given options is 9.6 m/s and that is, thus, the correct option.

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Free Fall!

2 solutions

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