Proof of 5.875m Initial Height of Batted Ball

[aisha]

h=height in m, of a batted baseball as a function of time t seconds since ball was hit

h=-4.5t^2+13.5t+5.875

can someone tell me what is the physical significance of the 5.875 and prove it? I am thinking maybe this is the height at which the ball was hit but if this is true then how will I prove it ? Please help all these quadratic questions are hurting my head

 

[check]

You're on the right track. To prove it, you know that time starts when the ball is hit (t = 0), so what is the height of the ball at this time?

 

[aisha]

Sooo Check is this right?

You're on the right track. To prove it, you know that time starts when the ball is hit (t = 0), so what is the height of the ball at this time?

so check do i take the eqn given and sub in 0 for t?
h=-4.5(0)^2+13.5(0)+5.875
h=5.875m?
thats it this then prooves that at 0 sec when the ball is hit the height of the ball is 5.875m

Hey thanks sooo much !

 

[ek]

so check do i take the eqn given and sub in 0 for t?
h=-4.5(0)^2+13.5(0)+5.875
h=5.875m?
thats it this then prooves that at 0 sec when the ball is hit the height of the ball is 5.875m

Hey thanks sooo much !

Make sure you denote that h(0) = 5.875, not just h.

Picky, yes, but still important.

 

[BobG]

h=height in m, of a batted baseball as a function of time t seconds since ball was hit

h=-4.5t^2+13.5t+5.875

can someone tell me what is the physical significance of the 5.875 and prove it? I am thinking maybe this is the height at which the ball was hit but if this is true then how will I prove it ? Please help all these quadratic questions are hurting my head

Are you sure you read the question right and put in the right numbers (and in the right places)?

Acceleration of gravity is actually closer to -4.9 m/s^2. And who's the batter, Joe Ascue?

The pitch is 5.875 meters high (about 19 feet high). It might be more likely that 5.875 m was the maximum height and that the question is asking how long the ball stayed in the air (that way you could look at the horizontal component of velocity and see how far the ball went). Note: To solve, you'll still move the h over to the right side of equation, so you can set the entire equation to zero, but then the height will be negative.

Edit: After thinking about it, I'll bet the wall the batter is trying to hit the ball over is 5.875 m high. Your final height is 5.875 m. If you set the equation to zero, it still means the batter's initial height was 5.875 meters below his target, so it should still be (-5.875) in your equation.
And, for those curious, Joe Ascue of the Indians played a key role in the Red Sox winning the 1967 pennant. Late in a 1-run game, the Red Sox tried to intentionally walk him with a man on third and one out, but Ascue outsmarted them. He grounded into a game ending double play.

 

[HallsofIvy]

"Acceleration of gravity is actually closer to -4.9 m/s^2. "

Actually, the acceleration of gravity is -9.8 m/s^2. The "-4.9" is g/2.

 

[aisha]

Are you sure you read the question right and put in the right numbers (and in the right places)?

Acceleration of gravity is actually closer to -4.9 m/s^2. And who's the batter, Joe Ascue?

The pitch is 5.875 meters high (about 19 feet high). It might be more likely that 5.875 m was the maximum height and that the question is asking how long the ball stayed in the air (that way you could look at the horizontal component of velocity and see how far the ball went). Note: To solve, you'll still move the h over to the right side of equation, so you can set the entire equation to zero, but then the height will be negative.

Edit: After thinking about it, I'll bet the wall the batter is trying to hit the ball over is 5.875 m high. Your final height is 5.875 m. If you set the equation to zero, it still means the batter's initial height was 5.875 meters below his target, so it should still be (-5.875) in your equation.
And, for those curious, Joe Ascue of the Indians played a key role in the Red Sox winning the 1967 pennant. Late in a 1-run game, the Red Sox tried to intentionally walk him with a man on third and one out, but Ascue outsmarted them. He grounded into a game ending double play.

well the equation was given to me by the teacher, so I can't change it to -5.875. I understand my answer is not right because it is not possible to hit a ball that high, but I already know the maximum height of the ball which is 16m at 1.5seconds this I got by completing the square, so the wall is 5.875? but the person hitting, hit it higher than the wall maximum of 16m? Is this the right answer? So then the way i was prooving it by subing in 0 for seconds is that still write ?

NOTE: There is no velocity involved, this is just a max/min quadratic problem.

 

[BobG]

Yes, 16 meters at 1.5 seconds and you solve it by setting the equation equal to zero.

The 16 meters is the maximum height of the ball. The path of the ball is a parabola and the max height occurs around mid path, right? So, the max height of the ball is only partly dependent on the height of the batter and the height of the wall.

Funny, he'd word the question this way, though. It's almost as if you're looking at the picture in reverse motion, watching the ball go from the wall back to the batter.

 

[aisha]

I emailed my teacher and said are these numbers realistic? she said well think of it this way he could be standing on a building lol, I was like ok that's interesting thanks for all ur help