Solving an Initial Speed Physics Problem with a Baseball Thrown from the Street

[thschica]

A baseball is seen to pass upward by a window 24 m above the street with a vertical speed of 9 m/s. The ball was thrown from the street.
(a) What was its initial speed?
m/s ( I keep getting 26 what equation am I supposed to use?)
(b) What altitude does it reach?
m (I don't know what equation I am supposed to use)
(c) How long ago was it thrown?
s (I need an equation because I got a negative number.) :yuck:
(d) After how many more seconds does it reach the street again?
s (Once again I am not sure what equation I am supposed to use!)

 

[Jameson]

What work have you done?

When trying to choose what kinematic equation to use, think about what information you know, and what you need to solve for.

You are given displacement and final velocity. And, you know the acceleration (hint: gravity).

So which equation do you need to use?

 

[Mozart]

From the question I am assuming that the balls velocity is 9 m/s at a height of 24 m. You want to find out the maximum height the ball reaches. So we need to find out the distance that must be added to the given height of 24 m.

I would use the formula 2as=vf^2-vi^2 to solve for the unknown distance (s) You now can find out the rest of the distance the ball will travel after reaching the point of 24 m.

Remember that the acceleration due to Earths gravity is 9.8 m/s/s, in this case the ball will have a negative acceleration on the way up. You can also figure out what the final velocity is going to end up being right?

When you find the TOTAL distance traveled by the ball go back to 2as=vf^2-vi^2 and substitute in all the information to solve for Vi (initial velocity)

There are probably a few different ways to tackle this problem. Mine might be a stupid way of thinking about it so try all kinds of things.

 

[HallsofIvy]

You should know that v(t)= at+ v₀: a is the acceleration, t the time elapsed, v₀ the initial speed, v(t) the speed at time t.

Also h(t)= -(a/2)t²+ v₀t+ h₀. a, t, v₀ are as above, h₀ the initial height.

(Of course, your book might use different letters but you should be able to recognize the formulas).

Since the acceleration is due to gravity, a= g= 9.8 m/s². Also it would be reasonable to take the "initial height" to be 0.

"A baseball is seen to pass upward by a window 24 m above the street with a vertical speed of 9 m/s."

So at that time, v(t)= -9.8t+ v₀= 9 and h(t)= -4.9t²+ v<sub>0[/sup]t= 24. You have two equations for the two unknowns t and v0. I might recommend solving -9.8t+ v0= 9 for v0 (as a function of t) and plugging that into -4.9t²+ v0t= 24 to get a quadratic equation in t. After you find t (There will be two solutions. Do you see why you want the smaller solution here?), put that into the first equation and solve for v0. Do "throw away" that value of t- that's also asked "(c) How long ago was it thrown?".

h(t)= -4.9t²+ v0t is quadratic. Do you know how to find the maximum (vertex) of a quadratic function? (Try "completing the square".)</sub>

 

[thschica]

I have all of the answers but B.Can someone help me with it?
The answers so far are a.)23.5 c.)1.5 d.)3.3

 

[curly_ebhc]

solving part b

Part b) is actually quite easy but you must remeber to add 24 meters (thx Mozart) to your final answer.

Lets look at what we are given:
v initial = 9 m/s
acc = -9.80 m/s^2 (gravity)
Now let's look at the things we must assume:
lets set initial height to zero so that we only need to solve for change in height.
so:
initial x=0
we also assume that the ball must stop for an instant at the top and has no velocity
so:
v final =0

Lets put all givens together:
v initial = 9 m/s
acc = -9.80 m/s^2 (gravity)
initial x=0
v final =0
x final =?

Now we see that we have v initial, v final, acc, x initail and we are solving for x final. We need to find the equation with all these things.
(It is the no t eqn. because you don't use time)
Plug and chug, but don't forget to add 24 m to the answer you got.

P.S. You don't have to take out the 24m if you just expand delta x. (xf-xi)

 

[thschica]

Thanks so much I just got it right!