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butalik [34]
3 years ago
8

You throw a small rock straight up from the edge of a highway bridge that crosses a river. the rock passes you on its way down,

9.00 s after it was thrown. what is the speed of the rock just before it reaches the water 29.0 m below the point where the rock left your hand? ignore air resistance.
Physics
2 answers:
MAXImum [283]3 years ago
6 0
Looking for t, the computation would be:
t = 2[initial vertical velocity] / g = 2u/g 
=> 9 = 2u/9.81 
or u = 9 x 9.81 / 2 = 44.16 m/s (upwards) 
conservation of energy: loss in potential energy(mgh) would be equal to gain in kinetic energy (½mv²) 
=> ½mv² - ½mu² = mg29
=> v² = 58(9.81) + 44.16^2
= 568.98 + 1950.1056 = 2519.0856 
getting the square root will give us the speed of the rock, which is:
=> v = √[2519.0856] ~= 50.19 m/s (directed downwards) 
lisov135 [29]3 years ago
4 0

Time of flight=\frac{2v}{g}

⇒9=\frac{2v}{9.81}

⇒v=\frac{9.81{\times}9}{2}

⇒v=44.145 ms^{-1}

As the rock passes you, the speed is 44.145ms^{-1}.

Now, rock's path from bridge to water is:

v^{2}-u^{2}=2gs

⇒v^{2}=(44.145)^{2}+2(9.81)(29)

⇒v^{2}=1948.7+568.98

⇒v^{2}=2517.68

⇒v=50.17ms^{-1}

Thus, rock's speed is 50.17ms^{-1} just before it reaches the water.

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