Question

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.

Answer 1

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) 

Answer 2

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.