Question

A rock is thrown vertically with a velocity of 20m/s from the edge of a bridge 42m above a river. How long does the rock stay in the air?

Answer 1

First Step is to find out how far the rock goes up before dropping, at Ymax V=0
At the point acceleration due to gravity takes over

So in order to find Ymax we have to find the time at which it starts to drop
Vfinal = 0 So...

V=Vo-at
0=Vo-9.8t

Solve for t and you have the time it continues to increase in height before dropping

That solves for the first half of the duration.


Second Step:

This is the second half of the duration where it starts to fall and it finally hits the ground meaning Yfinal = 0 but also remember your'e 42m above the river so you had a height advantage to begin with at 42m

Now you start at a height of 42m so Yo=42m

We plug into the equation:
Y-Yo=Vot-(1/2)at^2

Vo is actually zero in this case because it's dropping dropping from Ymax therefore v=0 at max height.

0-42=0-(1/2)at^2

Solving for t should give you the second half of the duration

Finally:

Add t1 and t2 together you get you're total time.

Answer 2

For the first part, when it is going up, why is the acceleration of gravity used?

Answer 3

Disregard the first 4 lines of text, I thought you asking something else and then corrected myself.

Answer 4

Gravity is used because gravity LOVES to keep you DOWN.

Without gravity being used if you jumped up you'd never come back down.

Answer 5

Is this how the diagram should be?