Projectile Motion from a given height

[qwerty101]

Homework Statement

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
v_x=2.82 m/s
v_y=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?)

Homework Equations

(y-y₀)=V_0yt+at²/2
v=v₀+at
a_c=v²/R
a_t=dv/dt

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?

 

[haruspex]

x(range)= 1.72m

Much more, surely.

-What are the radiuses of the baseball path curvature at max height and at the point of grounding? (What does this question even mean?)

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.

Anybody know what I am doing wrong?

Without seeing your working?

 

[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 v_xt which got me 1.72. How is that wrong?

 

[haruspex]

Well to get the y max I first calculated time, which would be equal to -2v₀/g

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?

 

[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?

 

[haruspex]

To get range, I then multiplied the time by vxt=3cos20(.55) and got 1.55 m. Sound about right?

Yes. (Seemed too low until I realized that 3m/s would be a wet kipper of a bunt.)

I am now trying to calculate the speed of the ball right before it hits the ground using equations like v=v0+at

Again, show your working. (You will need to calculate the x and y components separately, then combine them into a single speed.)

 

[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?

 

[haruspex]

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?

Yes.

 

[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, I've tried that max height would occur at the time when v_y=0, so I used the equation v_y=v₀sinβ-gt and set v_y 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.

 

[haruspex]

So I've been trying to calculate the y max height. I thought that would occur at .5t, which would give me .275 s.

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?

 

[qwerty101]

um no I don't have that equation but that's what I was sort of talking about in the second part of my reply
"I've tried that max height would occur at the time when v_y=0, so I used the equation v_y=v₀sinβ-gt and set v_y 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."

 

[haruspex]

"I've tried that max height would occur at the time when v_y=0, so I used the equation v_y=v₀sinβ-gt and set v_y 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."

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

 

[qwerty101]

Thanks I have a question regarding centripetal motion... I know a_cp=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 v_x=v₀cosβ. But now I have no idea how to calculate the radius. Any suggestions sir/mam?

 

[haruspex]

Thanks I have a question regarding centripetal motion... I know a_cp=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 v_x=v₀cosβ. But now I have no idea how to calculate the radius.

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