SOLUTION: hello....I am stumped...hope you can help...Question: Assume Joe Ball hits a major league pop-up (straight upward) on a Johnny Bench 90mph pitch. The function that describes the ba

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: hello....I am stumped...hope you can help...Question: Assume Joe Ball hits a major league pop-up (straight upward) on a Johnny Bench 90mph pitch. The function that describes the ba      Log On


   

Question 139518: hello....I am stumped...hope you can help...Question: Assume Joe Ball hits a major league pop-up (straight upward) on a Johnny Bench 90mph pitch. The function that describes the ball's height (in feet/second) is: h(t) = -16t^2+132t h(t) is height, t = is seconds...How many seconds for the ball to go up and back down? How high does the ball go? I came up with answers 4.125 for ball to go up and down and 272.25 feet for how high ball goes up (correct??)...but here's the stumper: it's now asking to repeat problem using planet Mercury. How do I complete this? (FYI, Mercury is 38% of Earth)....Hope you can help and show me the correct formula and answers..Thank you
Found 2 solutions by scott8148, Earlsdon:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the 4.125 is time up and time down, so "time of flight" is 8.25

on Mercury, the gravitational acceleration (twice the coefficient of the x^2 term) is 38% of Earth
__ so just replace the -16 with (38%)(-16)



FYI: (the real) Johnny Bench was a catcher, not a pitcher

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let's look at part I of the problem:
I)
a) How many seconds for the ball to go up and down?
In other words, how many seconds for the ball to return to the ground?
Starting with the given formula:
h%28t%29+=+-16t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D where:
v%5B0%5D+=+132ft/sec This is the initial upward velocity.
h%5B0%5D+=+0 This is the initial height.
So we have:
h%28t%29+=+-16t%5E2%2B132t and you want to find the time, t, when h = 0:
0+=+-16t%5E2%2B132t Factor a t from the right-hand side.
0+=+t%28-16t%2B132%29 Apply the zero product rule: If a%2Ab+=+0 then either a+=+0 or b+=+0 or both.
Add 16t to both sides. So...
t+=+0 or -16t%2B132+=+0
-16t+%2B+132+=+0 Add 16t to both sides.
132+=+16t Divide both sides by 16.
t+=+8.25seconds.
The ball returns to the ground in 8.25 seconds.
b) How high does the ball go?
You'll need to find the value of t at the vertex of the parabola described by the given equation. This is given by:
t+=+-b%2F2a The a and b come from: ax%5E2%2Bbx%2Bc+=+0 and in this case, a = -16 and b = 132.
t+=+%28-132%29%2F2%28-16%29
t+=+4.125 seconds. This is the time at which the ball reaches its maximum height. Substitute this value of t into the given equation and solve for h to find the maximum height.
h%284.125%29+=+-16%284.125%29%5E2%2B132%284.125%29
h%284.125%29+=+-272.5%2B544.5
h%284.125%29+=+272.25 feet. Which is precisely what you got.
II) For this problem to be solved for the planet Mercury whose gravitational force is 38% that of earth's, let's look at the original formula for the height of an object propelled upwards:
h%28t%29+-+%281%2F2%29gt%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D
Here, the acceleration due to gravity is: g = 32 ft/second squared for earth.
For Mercury, g = 0.38(32)ft/second squared = 12.16 feet/second squared.
So the formula for the planet Mercury becomes:
h%28t%29+=+-%281%2F2%2912.16t%5E2%2Bv%5B0%5D%2Bh%5B0%5D
h%28t%29+=+-6.08t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D This is the formula you would use to solve the second part of the problem.
a)
h%28t%29+=+-6.08t%5E2%2B132t Set h = 0 and solve for t.
0+=+-6.08t%5E2%2B132t So...
t+=+0 or...
t+=+132%2F6.08
t+=+21.71 seconds. This is the time at which the ball returns to the ground on Mercury.
The time to reach its maximum height is:
t+=+-b%2F2a where: a = -6.06 and b = 132.
t+=+-%28132%2F2%28-6.08%29%29
t+=+10.86seconds. Substitute this into the formula for Mercury:
h%2810.86%29+=+-6.08%2810.86%29%5E2%2B132%2810.86%29
h%2810.86%29+=+-66%2B1433.52
h%2810.86%29+=+1367.52feet. This is the maximum height attained by the ball on Mercury.