Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
Intro. Prob. Set 4 #16
At a certain instant an object is released from rest into free fall, and a second object is struck, giving it a speed of 10.425 meters/second in the horizontal direction. The second object travels in a straight line at this speed.
After falling precisely .75 meters, the first object strikes the second object.
How far did the second object move in the horizontal direction?
Solution
To find the distance moved by the first object we need only find the time, and multiply by its velocity.
The time will be the time of fall of the second object, so we momentarily forget about the first object and determine how long the second object requires to fall.
We will use `dsy, v0y, vfy and ay for the displacement, initial and final velocities and acceleration in the vertical, or y, direction; `dsx, v0x, vfx and ax will stand for the corresponding quantities in the horizontal, or x, direction.
To find `dt from the known vertical quantities we use the formula `dsy = v0 * `dt + .5 ay `dt ^ 2, with v0=0, ay=9.8 m/s/s and `dsy = .75 m.
We obtain `ds = .5 a `dt^2, since v0 = 0.
Solving for `dt and rejecting the negative solution we get `dt = `sqrt( 2 * `ds / a ).
Substituting we obtain `dt = `sqrt{( .75 m/s)/4.9 m/s/s)} = .391 s.
The distance traveled by the first object in this time is
`dsx = ( 10.425 m/s)( .391 s) = 4.076175 m.
Generalized Solution
An object will fall freely a distance `dsy from rest in time `dt = `sqrt( 2 `dsy / g ), obtained from `dsy = v0y * `dt + .5 ay `dt^2 with v0 = 0 and ay = g.
In this time an object moving at a constant horizontal velocity vx will travel distance
`dsx = vx * `dt = vx * `sqrt( 2 * `dsy / g).
** **
Why is it that we use the same amount of time? I'm picturing a ball falling and striking another with the second then traveling horizontally. We found the time required for the first object to travel vertically .75 m. Why will the second object only travel horizontally for that same amount of time?
** **
As it travels in the horizontal direction, the second object also falls under the influence of gravity. It started its fall while moving only in the horizontal direction, so for the 'falling' phase of its motion its initial vertical velocity was zero, the same as that of the first ball.
Thus both balls reach the ground at the identical instant.
Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
Set 4 Problem number 14
--------------------------------------------------------------------------------
Problem
Solution
Generalized Solution
Problem
How long will it take an object to fall 7.2 meters if it is released into free fall from a state of rest?
Solution
In the vicinity of Earth's surface, the object will accelerate at 9.8 m/s/s.
If we were given the time duration and acceleration we could reason this problem out in two steps; but with the given information there is no easy way to reason out the time.
So we use the formula `ds = v0 * `dt + .5 a `dt ^ 2, with v0=0, a=9.8 m/s/s and `ds= 7.2 m.
Since v0 = 0 the formula becomes `ds = .5 a `dt^2.
Solving for `dt we obtain `dt = `sqrt{2 `ds / a}
Note that -`sqrt{2 `ds / a} is also a possible solution, but in this situation we reject the negative `dt as making no sense.
Substituting we obtain
`dt = `sqrt(2 * 7.2 m / (9.8 m/s^2) ) = 1.212 s.
Generalized Solution
Starting from rest and accelerating at constant rate g we have
`ds = v0 `dt + .5 a `dt^2 = .5 a `dt^2, since v0 = 0.
We solve `ds = .5 a `dt^2 for `dt and choose the positive value of `dt:
`dt = `sqrt ( 2 `ds / a ).
After falling precisely .75 meters, the first object strikes the second object.
How far did the second object move in the horizontal direction?
Solution
To find the distance moved by the first object we need only find the time, and multiply by its velocity.
The time will be the time of fall of the second object, so we momentarily forget about the first object and determine how long the second object requires to fall.
We will use `dsy, v0y, vfy and ay for the displacement, initial and final velocities and acceleration in the vertical, or y, direction; `dsx, v0x, vfx and ax will stand for the corresponding quantities in the horizontal, or x, direction.
To find `dt from the known vertical quantities we use the formula `dsy = v0 * `dt + .5 ay `dt ^ 2, with v0=0, ay=9.8 m/s/s and `dsy = .75 m.
We obtain `ds = .5 a `dt^2, since v0 = 0.
Solving for `dt and rejecting the negative solution we get `dt = `sqrt( 2 * `ds / a ).
Substituting we obtain `dt = `sqrt{( .75 m/s)/4.9 m/s/s)} = .391 s.
The distance traveled by the first object in this time is
`dsx = ( 10.425 m/s)( .391 s) = 4.076175 m.
Generalized Solution
An object will fall freely a distance `dsy from rest in time `dt = `sqrt( 2 `dsy / g ), obtained from `dsy = v0y * `dt + .5 ay `dt^2 with v0 = 0 and ay = g.
In this time an object moving at a constant horizontal velocity vx will travel distance
`dsx = vx * `dt = vx * `sqrt( 2 * `dsy / g).
** **
I initially used vf^2 = v0^2 + 2a`ds. I then had vf^2 = 0 + 2(9.8 m/s^2)(7.2 m). vf^2 = 141.12 m^2/s^2. vf = 11.88 m/s. This is not the correct final velocity. Why?
** **
I don't believe the final velocity was calculated in the given problem.
The final velocity you obtained is consistent with the given solution, in which the time of fall was found to be about 1.2 seconds.
Using your final velocity the average velocity would be (0 + 11.88 m/s) / 2 = 5.94 m/s. Multiplying this by 1.2 seconds gives us about 7.2 meters, which is the distance of fall.