A log is floating on swiftly moving water. A stone is dropped from rest from a 50.8-m-high bridge and lands on the log as it passes under the bridge. If the log moves with a constant speed of 5.36 m/s, what is the horizontal distance between the log and the bridge when the stone is released?

Horizontal distance between the log and the bridge when the stone is released = 17.24 m

Step-by-step explanation:

Height of bridge, h = 50.8 m

Speed of log = 5.36 m/s

We need to find the horizontal distance between the log and the bridge when the stone is released, for that first we need to find time taken by the stone to reach on top of log,

We have equation of motion. s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 50.8 m

Substituting,

s = ut + 0.5 at²

50.8 = 0.5 x 9.81 x t²

t = 3.22 seconds,

So log travels 3.22 seconds at a speed of 5.36 m/s after the release of stone,

We have equation of motion. s = ut + 0.5 at²

Initial velocity, u = 5.36 m/s

Acceleration, a = 0 m/s²

Time, t = 3.22 s

Substituting,

s = ut + 0.5 at²

s = 5.36 x 3.22 + 0.5 x 0 x 3.22²

s = 17.24 m

Horizontal distance between the log and the bridge when the stone is released = 17.24 m