A woman on a bridge 82.2 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 5.04 m more to travel before passing under the bridge. The stone hits the water 2.13 m in front of the raft. Find the speed of the raft.

0.71 m/s

Explanation:

We find the time it takes the stone to hit the water.

Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.

So, substituting the values of the variables into the equation, we have

y = ut - 1/2gt²

82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²

82.2 m = 0 + (4.9 m/s²)t²

82.2 m =  (4.9 m/s²)t²

t²  = 82.2 m/4.9 m/s²

t² = 16.78 s²

t = √16.78 s²

t = 4.1 s

This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.

Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.