Help with projectile motion problem Solution?

In summary, the home run was hit at a very high angle and with negligible air resistance. The ball reached the wall in 4.31 seconds.
  • #1
kudelko24
11
0

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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  • #2
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?
 
  • #3
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
 
  • #4
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!
 
  • #5
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
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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!
 

Related to Help with projectile motion problem Solution?

1. What is projectile motion and why is it important?

Projectile motion refers to the motion of an object through the air, under the influence of gravity. It is important because it helps us understand how objects move and interact with their surroundings, and it has many practical applications such as in sports, engineering, and physics experiments.

2. How do I solve a projectile motion problem?

To solve a projectile motion problem, you can use the equations of motion, which relate an object's initial velocity, acceleration, and time to its final position. You will also need to use trigonometry to calculate the horizontal and vertical components of the object's motion.

3. What are the key factors that affect projectile motion?

The key factors that affect projectile motion are the initial velocity, angle of projection, air resistance, and the force of gravity. These factors determine the trajectory, distance, and time of flight of the object.

4. Can you provide an example of a projectile motion problem?

Sure, an example of a projectile motion problem could be: A football is kicked with an initial velocity of 20 m/s at an angle of 45 degrees to the ground. How far will the ball travel and how long will it take to hit the ground?

5. How can I check if my solution to a projectile motion problem is correct?

You can check if your solution to a projectile motion problem is correct by using the laws of physics, such as the conservation of energy and momentum, to verify that the calculations are accurate. You can also use online calculators or simulation tools to compare your results.

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