2 Two-dimensional motion problems

[MJay82]

I spent all day yesterday trying to figure out how to do these on my own, but I can't seem to conceptualize these two problems.

Problem 1

Homework Statement

Background
When baseball outfielders throw the ball, they usually allow it to take one bounce on the theory that the ball arrives sooner this way. Suppose that after the bounce the ball rebounds at the same angle θ as it had when released (see figure below) but loses half its speed.

Problem
Assuming the ball is always thrown with the same initial speed, at what angle θ should the ball be thrown in order to go the same distance D with one bounce as one thrown upward at ϕ = 43.8° with no bounce?

Homework Equations

My teacher said I needed to use this trig identity:
sin(2u) = 2sin(u)cos(u)

The Attempt at a Solution

I attemped this problem in this manner.
R(1bounce) = (Vi sin2R2R(1 bounce) = R(0 bounce)
sin(2\Phi)=2sin(\Theta)cos(\Theta)
.4996=sin(\Theta)cos(\Theta)This doesn't seem right. Or at least I don't know the trig identity that can get me \Theta from here.For this second problem, I figured out the initial height and the range, but I don't know how to work these into a problem that can get the time I need at part C.

Homework Statement

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.50 m/s at an angle of 16.0° below the horizontal. It strikes the ground 6.00 s later.

(a) How far horizontally from the base of the building does the ball strike the ground? (Xf)
(b) Find the height from which the ball was thrown. (Yi)
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Xi = 0 m, Xf = range = 43.26 m
Yi = 188 m, Yi = 0 m
Vxi = 7.21 m/s, Vyi = -2.07 m/s
Vxf = 7.21 m/s, Vyf = -2.07 m/s - 9.8t
Ax = 0 m/s^2, Ay = -9.8 m/s^2
t = 6.00 s

Homework Equations

Kinematic equations

The Attempt at a Solution

I initially thought this part (part c) was easy - I just used Yf = Yi + (Vyi)t + (a/2)t^2.

Obviously I didn't account for the change in displacement of X, and treated the problem as if the object was dropped straight down. So I'm stuck, and don't know where to go from here. Any help is greatly appreciated.

 

[Delphi51]

Problem 1 is most interesting; thanks for posting it.
I don't see how you did it. Surely you need to work out the vertical part of the motion to get the time of flight? Then the horizontal distance of the flight. I don't see any time or distance calcs in your work.

You can solve .4996 = sin(θ)*cos(θ) by using that identity:
2*.4996 = 2sin(θ)*cos(θ)
2*.4996 = sin(2θ)
I see that gives θ = 43.8, but of course that is not the answer to the question. You would have to throw at a smaller angle so the ball hits the ground after a smaller distance, leaving room for the bounce.

 

[MJay82]

Thanks for the reply, Delphi. This assignment was due last night. I actually ended up getting the first question correct by creating a table of answers to problems similar to this one. Not the way I would have liked to solve it, but I'll take a 90 over a 70 and learn how to do it next week. The answer was 26.5 degrees.

 

[Delphi51]

For next week . . . nearly all projectile motion questions yield to this elegant method:
- make headings for horizontal and vertical parts of the motion
(the two parts are independent; one does not affect the other)
- there are no horizontal forces, the motion is not accelerated so you write d = vt or perhaps x = Vx*t if you are picky.
- under the vertical heading, you write d = Vi*t + ½at² and
v = Vi + at (though this last formula is not needed in this particular question)
- fill in the things you know. For the horizontal part, I wrote
147 = v*cos(37)*t (where v is the initial speed of the ball).
This gives t = 147/(v*cos(37)). [1]
In the vertical part, you get
26 = v*sin(37)*t - ½gt² [2]
The classic two equations with two unknowns. Sub 1 into 2 and you soon get a number for the initial velocity. And the time of flight if you want it.

The next step is to use that initial velocity for the throw with a bounce. This time you use the velocity equation for the vertical part and not the distance one, just because the vertical distance is neither known nor needed.

Don't let this one go by . . . it is a very widely needed technique!