If a soccer ball is kicked straight up from the ground with an inital velocity of 32 feet pre second...?

Question

then its height above tha earth in feet ia given by s(t)=-16t^2+32t where t is time in seconds. Graph this parabola for 0(less than or equal to)t(less than or equal to)2.What is the maximum height reached by the ball?

Answer 1

for max height take derivative of s
-32t+32=0

for t=1
s(t) is max

s(1)=-16+32=16
16 is the max height reached by the ball.

To graph s(t) between 0 and 2,
find x-intercept and y-intercept and vertex of the parabola
The arms of the parabola is down because of the minus sign in front of t^2
Try to draw the graph by yourself. It is too easy.

Answer 2

If the velocity is 32 per second, and the gravity has a pull of 16 per second, the ball will only travel for 1 second upwards where it will reach and peak at 16 feet, considering that the the ball reaches 0 again at 2 seconds.
s(t) = -16t^2 + 32t
x = -b/2a (vertex formula)
x = -32/2(-16)
x = -32/32
x = 1

Answer 3

Negating the effects of wind resistance (which we commonly do in physics class), the formula to use would be Vf^2=Vo^2+2AS.

Vf (final velocity) in this case is zero, since we want to find the maximum height of the ball. Its velocity will be zero because at the transition point between upward movement and downward movement (the maximum height) the ball literally stands still (has zero velocity) for a split second.

Vo, then, is the initial velocity, or 32 ft/sec as given in the problem.

A is the constant acceleration of gravity (-32 ft/sec^2).

S, is the unknown, or the height of the ball at its zero-velocity point.

Solving for S, we get S=(Vf^2-Vo^2)/(2A), or S=(0-(32^2))/(2*(-32)), which is 16ft.

Consequently, this is the same answer you get when you solve the equation given in your question at a time of t=1second.

The graph of this equation is an inverted parabola, with time (t) on the x axis going from 0 to 2 seconds. The y axis is height (s), goes from zero to 16 feet. At a time of t=0, the ball is at 0ft. At a time of t=1sec, the ball is at its peak, or 16ft high. At a time of t=2sec, the ball is back on the ground at 0ft.

The shape of the graph looks like this (if yahoo displays it right):

*****************#*****************
**********#*************#**********
******#*********************#******
***#***************************#***
*#*******************************#*
#*********************************#

(Well yahoo didn't display it right, but follow the contour of the graph back down to zero. You get the idea.)

Answer 4

Maximum height is 16 feet from ground.

You take the derivative and set it to Zero to find the maximum height.
The derivative gives the slope, and when the slope is zero, you're at the vertex of the parabola.

s'(t) = -32t + 32 = 0

t = 1 s'(1) = 0

s(1) = 16

The graph of the parabola is downward at (t=1, s(t)=16), and it crosses the x axis at t=0, and t=2.

Answer 5

Maximum height is when t = 1
giving S = 16 feet.