# Homework Help: Projectile Motion from a given height

1. Feb 7, 2013

### qwerty101

1. The problem statement, all variables and given/known data
A baseball is bunted down the third base line leaving the bat from a height of .900m with initial v = 3.00 m/s at angle 20° above the horizontal.
I calculated that
x(range)= 1.72m
t(total)=.61s
vx=2.82 m/s
vy=1.03 m/s
-What is the speed of the ball before it hits the ground?
-What is the max height of the baseball above the horizontal in this travel?
-What is the ball's centripetal and tangential acceleration at max height?
-What is the ball's centripetal and tangential acceleration at the instant just before the ball hits the ground?
-What are the radiuses of the baseball path curvature at max height and at the point of grounding? (What does this question even mean?)

2. Relevant equations
(y-y0)=V0yt+at2/2
v=v0+at
ac=v2/R
at=dv/dt

3. The attempt at a solution
For calculating the max height, I keep getting y max as equal to .478 m, which makes no sense since its even less than the initial height... Anybody know what I am doing wrong?
Additionally for for the speed before it hits the ground I'm pretty sure that means normal velocity, yet when I plug in the numbers to the equation I get a velocity of -2.98 m/s which really does not make sense.
Regarding the centripetal acceleration, I just wanted to clarify if r(radius) would be equal to half of the distance traveled or not...
And how to calculate tangential acceleration? I have no clue...
Also what would the difference between the acceleration at max height and before it lands?

2. Feb 7, 2013

### haruspex

Much more, surely.
Do you know about curvature as a local property of a curve? Just as you can have a tangent line at a point on a curve, you can also have a tangent circle, which fits both the slope and curvature of the curve at that point.
Without seeing your working?

3. Feb 7, 2013

### qwerty101

Well to get the y max I first calculated time, which would be equal to -2v0/g, which got me .61 s. I then multiplied vxt which got me 1.72. How is that wrong?

4. Feb 7, 2013

### haruspex

That's the time to do what, exactly? Remember that the ball is struck while 0.9m off the ground. And what velocity are you using in that equation?

5. Feb 7, 2013

### qwerty101

The time it would take the ball to travel, but I realized the error in calculating time. I now used the equation y0-y=v0(sinβ)t-.5(g)t2 and got .55 s for t. To get range, I then multiplied the time by vxt=3cos20(.55) and got 1.55 m. Sound about right? I am now trying to calculate the speed of the ball right before it hits the ground using equations like v=v0+at and get completely different answers. Is the speed calculated supposed to be in a specific dimension or something?

6. Feb 7, 2013

### haruspex

Yes. (Seemed too low until I realised that 3m/s would be a wet kipper of a bunt.)
Again, show your working. (You will need to calculate the x and y components separately, then combine them into a single speed.)

7. Feb 7, 2013

### qwerty101

well i got the velocity x component by mulitplying initial velocity by cos(20) and got the y component by multiplying initial velocity by sin20 and then subtracting it by g(time). wait.... do I take the sqrt of these added and squared? just remembered..... so sqrt (vx2+vy2) right?

8. Feb 7, 2013

Yes.

9. Feb 7, 2013

### qwerty101

So I've been trying to calculate the y max height. I thought that would occur at .5t, which would give me .275 s. I then plugged this into the equation y-y0=(v0sinβ)t -.5gt2 so y-y0=-.088 and then added .9 which is y0. But this yields a number that is lower than the initial height, so makes no sense. NExt, ive tried that max height would occur at the time when vy=0, so I used the equation vy=v0sinβ-gt and set vy equal to zero to determine the time when the max was reached. For time I then got .1047 s. I then used the equation y-y0=(v0sinβ)t -.5gt2 again and got .9537 m. Does this sound about right? I'm not quite sure as it seems very small.

Last edited: Feb 7, 2013
10. Feb 7, 2013

### haruspex

You meant .55s/2, I assume. But .55s was the time to hit the ground, not the time to return to the initial height. Anyway, there's a simple way to get max height. It's the point where vertical velocity is zero. Do you have an equation like v2-u2=2as?

11. Feb 7, 2013

### qwerty101

um no I don't have that equation but thats what I was sorta talking about in the second part of my reply
"I've tried that max height would occur at the time when vy=0, so I used the equation vy=v0sinβ-gt and set vy equal to zero to determine the time when the max was reached. For time I then got .1047 s. I then used the equation y-y0=(v0sinβ)t -.5gt2 again and got .9537 m. Does this sound about right? I'm not quite sure as it seems very small."

12. Feb 7, 2013

### haruspex

Yes, that's about right. The equation I referred to is quite useful. It's basically conservation of energy with the mass term cancelled out: mv2/2 = mu2/s + mas, where u = initial speed and v = final speed, s = distance.

13. Feb 7, 2013

### qwerty101

Thanks I have a question regarding centripetal motion.... I know acp=v2/R. So to to find it during ymax, I first started by finding the velocity it would be, which would be the horizontal component since the vertical component is 0, which got me 2.819 m/s (by solving vx=v0cosβ. But now I have no idea how to calculate the radius. Any suggestions sir/mam?

14. Feb 7, 2013

### haruspex

You know the velocity and the acceleration, so you can calculate the radius using you formula.