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Question

I need help with this algebra problem...i have no idea how to solve it.........Kim throws a ball straight up into the air. The ball reaches a maximum height of 8 feet above the ground in .5 seconds.
a) Find the equation giving the ball's height (h) as a function of time (t).
b) When does the ball hit the ground?

Answer 1

You really need to be provided some background information for this type of problem. I assume you were given this.

Part a) requires you to know that the height of the ball is a quadratic of the form h=-16t^2 + vi * t + si where si is the initial height (Kim's height) and vi is the initial velocity or speed. This formula was probably given to you somewhere.

Since this is a quadritic, the maximum value is the vertex. The vertex x-coordinate is -b/(2a). For our equation the x-coordinate of the vertex is .5, so

-vi/(2*-16)=.5,

so vi is 16. And we know that at this vertex, the height is 8, so

8=-16(.5)^2 + 16*(.5) +si

so si=4

Now you have the complete funtion: h = -16*t^2 + 16*t + 4

To find when it hits the ground, just solve for h=0. You can do that part, right? Just put it in the quadratic formula. t=-2+2*sqrt(2)

(euclid jr is wrong)

Answer 2

Hi,

Using the fact that gravity has the effect of -16 feet/sec, you can write the equation representing the movement of a ball thrown upward by the equation y = -16( x - .5)^2 + 8. This equation is in the vertex form y = a(x - h)^2 + k, where (h,k) is the vertex, in your case the highest point.

If you put the formula into a graphing calculator and find the zero, it occurs at about 1.207 seconds as the time it hits the ground. If you don't have a graphing calculator, simplify the equation and get y = -16x^2 + 16x + 4. Then use the quadratic formula to solve it. With the help of a square root on a calculator, you can find this answer.

Answer 3

Solution:

a. given that t = 0.5 when h = 8, we can reasonably assume that h is a function of t by some constant, a. So,

t = a*h

now plugging in the given numbers into our equation, we get:

1/2 = a*8

a = 1/16

so the function is t(h) = h/16

b. Since 8 feet is it's max height, to hit the ground, the ball must travel another 8 feet, for a total of 16 - 8 feet to go up, 8 feet to go back down. Using our function, we calculate the value of t as:

t(h) = h/16

t = 16/16 = 1

So the ball will hit the ground in 1 second.

Answer 4

First part
Take negative accelaration, gravity=10m/s2
Initial velocity = u
final velocity = v
time, t = 0.5seconds

height = h

now we know that

final velocity = initial velocity + a production of accelaration and gravity

or v= u + at
Substituting values
0 = u+ (-10 x 0.5)
solving you get
u = 5 m/s


Using newton's law of motion


h= ut + (at2)/2
h = 5t + (-10 x t2)/2
h = 5t - 5t2, so here is your equation

second part
as for the ball hitting the ground

now yr accelaration is positive, and initial velocity is zero

so u again use

h = ut + at2/2

h= 8m, as given in the problem

so 8 = 0 + 10 x t2/2
8 x 2/10 = t2

t2= 8/5, solve that and u have yr answer

Answer 5

I don't need to answer this problem but my advice is ,you read your physics book or if not ask assistance from you teacher