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Baseball throw

  • #1
1. A third baseman makes a throw to first base 38.6 m away. The ball leaves his hand with a speed of 38.0 m/s at a height of 1.4 m from the ground and making an angle of 24.2 degrees with the horizontal. How high will the ball be when it gets to first base?



2. d=V0(t) + 1/2(g)(t^2)



3. 38.6= 38(t) + 1/2(-9.81)(t^2) using the quadratic equation I get 1.2 s for t. I'm not sure what to do from here.
 

Answers and Replies

  • #2
133
0
Remember to break the motion into its horizontal and vertical components. The equation you listed in #2 is valid for the vertical motion, yet in #3 you have plugged in the horizontal distance which is mixing apples and oranges.

What is the appropriate formula #2 for the horizontal motion?
 
  • #3
For the horizontal component, are you referring to d=Vo(cos24.2)(t)? Sorry I'm pretty confused with the formulas for projectile motion equations.
 
  • #4
133
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Yeah, that's it. Since there is no force on the horizontal (gravity is straight down), the basic formula is [itex]x = vt[/itex] which can be seen as a special case of [itex]x = v_0 t + \frac12 at^2[/itex] with [itex]a = 0[/itex].

Now, you know that v0 = 38.0 m/s. The cosine is there because this initial velocity is actually at an angle. Since we are focusing now on the horizontal motion, we only want the horizontal component of this velocity = (38.0)(cos 24.2).

You also know the horizontal distance. This is the left-hand side of the equation. The last variable is time -- which you can now solve for. This is the amount of time the ball is in the air.

This is as much information as you can squeeze from the horizontal motion. Now consider the vertical motion.
 
  • #5
Ok, so the x-component is 38(cos 24.2)=34.7 and the y-component is 38(sin 24.2)=15.6

And 38.6=34.7t which = 3.9s. Now how to go further?
 
  • #6
133
0
The formula you need is [itex]y = v_{0y} t + \frac12 a_y t^2[/itex]. You are looking for [itex]y[/itex] (how far does the ball drop in height). You now know [itex]v_{0y}[/itex], [itex]t[/itex] and [itex]a_y = -9.8 \text{ m/s}[/itex].
 
  • #7
I finally figured this one out after extensively reading the text book. I used the formula you just posted, and used the quadratic forumla to find the time. I then used Vo(cos24.2)t to get the distance, which I used to figure the 12.8m answer.

I'm having trouble with another projectile-type problem, and I'm going to work it some more. If I still can't figure it, I think I may just scan my work and post as an image for help.

I appreciate all of you help dulrich!
 

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