1. What is the vertical component of the ball's velocity right before Sarah catches it?
Question: what is the difference between this and the ball's velocity when Julie threw it?
Your intuition is correct that the velocity will be the same when Sarah catches it as when Julie first throws it, except, of course, for the fact that the vertical component will be aiming downward, rather than upward. When Julie threw the ball, the vertical component of the velocity was 24*sin(34) or around 13.421 m/s upward. Since Sarah catches the ball at the same height off the ground (1.5 m), the ball's vertical velocity will be -13.4 m/s (i.e. 13.421 m/s downward).
If there is a difference, how may i work on this problem?
If Sarah had been taller or shorter or had reached up or down to catch the ball, there are a couple of ways you could proceed. The most straightforward is probably just to use
,
where
is the height at which Sarah caught the ball (1.5m in this question, but it could be anything),
is the height where the ball started (i.e. the height from which Julie threw the ball, or 1.5 m),
is the initial vertical component of the velocity, 13.421 m/s,
is the acceleration due to gravity (-9.8 m/s^2). (Note that it's negative since gravity accelerates the ball
downward). Then we just solve for the time,
. This is a quadratic equation in
, so you'll probably need to use the quadratic formula to solve it. And bear in mind that means you will get
two solutions for
. The two solutions correspond to the two different times the ball would be at the specified final height during its parabolic flight -- one time
before reaching the apex of the trajectory and one time
after. You, of course, want the one after, since that's the one that corresponds to Sarah catching the ball. So you'll always choose whichever solution for
is greatest.
Once you know the total time required, you can use
to compute the vertical component of the velocity when Sarah catches the ball (since you know the values of all the variables on the right side of the equation).
Even though you can use your intuitive trick to easily calculate the vertical velocity like we did in the first paragraph, you should get the same answer if you do it the long way, as in the subsequent paragraphs. You may want to try it just to prove to yourself that it works out.
Finally, please remember that the
horizontal component of the velocity will be unaffected by gravity, time, distance, or anything else (ignoring air resistance of course). It just remains constant throughout the ball's flight. It was 24*cos(34), or around 19.897 m/s,when the ball left Julie's hand, and that's what it still is when Sarah catches it.
2. After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 14 m above the ground, and takes 2.592 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand?
Okay, so first of all we need to know how far apart Julie and Sarah are standing. Luckily we can get the answer from the information given in Part 1. You solved for the time, t, when Sarah caught the ball. You also know the horizontal component of the velocity was 24*cos(34) = 19.897 m/s. So now we can compute the horizontal distance between the two girls as
,
where
is the distance we're trying to find,
is the horizontal component of the velocity (19.897 m/s), and
is the time you computed in Part 1.
Now that you know how far apart they're standing, you can use the same formula to compute the horizontal component of the velocity of Sarah's extra-high throw because you know that it took 2.592 s to reach the point directly over Julie's head. So plug those two numbers (
and
) back into the above formula and solve for
.
Now we can compute the
vertical component of the ball's velocity because we know the maximum height it reached. The easy way to do this is to realize that the ball's upward velocity when it left Sarah's hand would be the same as the downward velocity of a ball dropped from 14 m to a height of 1.5 m. In that case, you'd solve the same equation from Part 1:
,
where x (the ending height) is 1.5 m,
is the starting height of 14 m,
is 0 since the ball would be starting from rest, and
would have course be -9.8 m/s^2. You'd solve this equation for
then plug that value back into
,
That would give you the value, v, of the vertical velocity of the ball when it reached 1.5 m. Just keep in mind that whatever velocity you calculate (positive or negative), the vertical velocity of the ball when it left Sarah's hand was positive (upward).
For what it's worth, if you didn't recognize the shortcut (that you can calculate the vertical velocity by pretending the ball was dropped from a height of 14 m), you could still use the same two formulas from above to calculate the vertical velocity directly. You'd use
as your initial velocity (for which you're trying to solve) and
as the velocity when the ball reaches 14 m (i.e. 0 m/s). The only complication is that you now have to simultaneously solve two equations and two unknowns (
and
). That's no big deal since it's trivial to solve the second equation for
in terms of
; then you'd just plug that back into the first equation in place of
and solve for
. Then plug that answer back into the second equation and you've got
. Either way you do it, you should get the same answer.
Once you know the horizontal and vertical components of the velocity, it's easy to compute the vector velocity:
where the
h and
v subscripts represent the horizontal and vertical components, respectively.
3. How high above the ground will the ball be when it gets to Julie?
I think when they say "when it gets to Julie", the mean "when it is directly over Julie's head".
Now that you know the initial vertical velocity when it left Sarah's hand, and you know the initial height (1.5 m), and you know the total time it took to get to Julie (2.592 s), you once again use the formula
,
to compute the position,
, of the ball when it reaches Julie.
I hope that helps!