Physics, 30.10.2020 18:10, lola06032003
A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.
Answers: 2
(Eq. 1)
, 
- Initial and final gravitational potential energies, measured in joules.
, 
- Initial and final translational kinetic energies, measured in joules. 
(Eq. 2)
- Gravitational acceleration, measured in meters per square second.
, 
- Initial and final height, measured in meters.
, 
- Initial and final speed of the rock, measured in meters per second. 
, 
, 
and 
, then the equation is:![\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-0\,m) = \frac{1}{2}\cdot \left[\left(29\,\frac{m}{s} \right)^{2}-v_{1}^{2}\right]](/tpl/images/0855/7188/55462.png)
, 
, then the equation is:![\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-y_{2}) = \frac{1}{2}\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(14.607\,\frac{m}{s} \right)^{2}\right]](/tpl/images/0855/7188/296c6.png)
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A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits t...
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