Help with projectile motion problem Solution?

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a baseball hit over a wall. The baseball must clear a height of 21 meters while traveling a horizontal distance of 130 meters, with the ball being hit at an angle of 35 degrees above the horizontal. The challenge is to determine the initial speed, time to reach the wall, and the velocity components at that point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion in both the x and y directions, questioning how to isolate variables such as initial velocity and time. There are attempts to express the initial velocity in terms of its components using trigonometric functions, and some participants express confusion about handling multiple unknowns in the equations.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between the variables involved. Some have suggested substituting expressions for time and initial velocity into the equations to reduce the number of unknowns. There is a mix of understanding and confusion, with some participants expressing a desire to clarify their reasoning further.

Contextual Notes

Participants note that the problem involves assumptions such as negligible air resistance and that the ball is hit from a height of 1 meter. There is also mention of reference answers from a textbook, which adds to the participants' motivation to understand the problem better.

kudelko24
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Homework Statement


A home run is hit in such a way that the baseball just clears a wall 21m high, located 130m from home plate. The ball is hit at such an angle of 35degrees to the horizontal, and air resistance is negligible. Find initial speed, the time it takes the ball to reach the wall, and the velocity components and the speed of the ball when it reaches the wall. (Assume that the ball is hit at a height of 1m above the ground. Suggestions or solutions please? I'm stumped!


Homework Equations


x=vo X t + .5 X a X t^squared


The Attempt at a Solution



?
 
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x(t)=v0_x*t (there is no acceleration in the x direction).
y(t)=y0+v0_y*t+(1/2)*g*t^2 (acceleration in the y direction is g downward).
v0_x and v0_y are related to the initial speed v0 by trig functions of the angle, right? I don't think they would give this question to a complete beginner. You must be able to do a better job of at least starting to work out the question. If t=0 is when the ball is hit. What is y0?
 
Help?

Ok... well, a^y = -9.8 m/s^2
theta = 35 degrees
y = 21 - 1 = 20 m, because of the ball being hit 1 m above the ground
x = 130 m

But you can't solve for the initial velocity or the horizontal/vertical components... how do i solve for the initial velocity? Or the time?

Vo = x / t
 
v0_x=v0*cos(theta), v0_y=v0*sin(theta). Haven't you ever seen anything like that before? Split the balls motion into x and y components!
 
ok... so the initial velocity in the x direction is Vox = Vo X cos (theta).
since you don't have the initial velocity it cancels out the cos (theta). Correct?
Same goes with y initial velocity.

Vo X .819 = Vox
Vo X .574 = Voy
 
Nothing is cancelling anything out. The numerical value of cos(theta)=0.819. That's ok. You don't have the initial velocity so you have to leave it in the equations as an unknown until you can solve for it in the end. Next, use the equation describing the x component of the motion to find the time to reach the fence. Your answer for t will have the unknown v0 in it.
 
ok well I'm confused... I have the answers in the back of the book to check my answers but that's not the problem. I want to know how to get those answers.

If the equation x = Vo x t, then 130 = Vo x t. I still have 2 unknowns in this problem. The initial velocity and the time. I can't solve for 1 if I don't have the other. Unless I am missing something here and completely oblivious to what is in front of me.

Here's the thing, if the answer is 3.8s and to find the initial velocity (Vo), I would take 130/3.8 = 34.2 m/s. But that is not the correct answer.
 
t=(130m)/(v0_x). That's the time when it reaches the fence. Yes, you still have two unknowns in the problem. That's ok, because we aren't done. Now write down the equation for the y motion.
 
y = Vyo X t - 1/2 a X t^2

20 = Vyo X t - .5(-10m/s)t^2

Unknown again is initial velocity and time
 
  • #10
Well, I'll try to work on this more this weekend. I'll be out of town until Sunday afternoon, so I'll take a look at it then. Thanks for all of your help. I'm determined to figure this problem out!
 
  • #11
Ok, so t=130m/(v0_x) is when it reaches the fence. Put that value of t into the y equation. Now there is only one unknown, v0.
 
  • #12
y = Vyo X t - 1/2 a X t^2

20 = Vyo X 130m/(v0_x) - .5(-10m/s)130m/(v0_x)^2
 
  • #13
kudelko24 said:
ok... so the initial velocity in the x direction is Vox = Vo X cos (theta).
since you don't have the initial velocity it cancels out the cos (theta). Correct?
Same goes with y initial velocity.

Vo X .819 = Vox
Vo X .574 = Voy

I wish you'd try a little harder. You got that stuff you wrote before. And t^2=(130m/v0x)^2. You could finish this anytime you want.
 
  • #14
dx=(Vix)(t)
130=(Vcos35)(-2(Vsin35)/-9.8)
130=V^2(cos35)(0.117)
V^2=130/0.096
V=36.82 m/s

t=(-2(Viy)/g)
t=(-2(Vsin35)/-9.8) or t=(-2(21.12 m/s)/-9.8)
t=4.31 s

Vy= Vsin35 = 21.12 m/s
Vx= Vcos35 = 30.16 m/s

I hope this helped!
 

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