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, 8.00 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.

Express your answer with the appropriate units

v=

Answer 1

time of flight = t, is defined as the time taken to reach high point of trajectory and return to the same initial horizontal level of projection point (here the bridge)

t = 2[initial vertical velocity] / g = 2u/g

=> 8 = 2u/9.81

or u = 8x 9.81 / 2 = 3.5 x 9.81 = 39.24 m/s (upwards)

conservation of energy: (neglecting air resistance frictional energy loss)

loss in potential energy(mgh) = gain in kinetic energy (½mv²)

=> ½mv² - ½mu² = mg29

=> v² = 58(9.81) + 39.24² = 568.98 + 1539.7776 = 2108.7576

=> v = √[2108.7576 ] ~= 45.92 m/s (directed downwards)