SOLUTION: `SUPPOSE YOU THROW A BASEBALL STRAIGHT UP AT A VELOCITY OF 64 FEET PER SECOND. A FUNCTION CAN BE CREATED BY EXPRESSING DISTANCE ABOVE GROUND, S, AS A FUNCTION OF TIME, T. THIS F

Algebra ->  Linear-equations -> SOLUTION: `SUPPOSE YOU THROW A BASEBALL STRAIGHT UP AT A VELOCITY OF 64 FEET PER SECOND. A FUNCTION CAN BE CREATED BY EXPRESSING DISTANCE ABOVE GROUND, S, AS A FUNCTION OF TIME, T. THIS F      Log On


   

Question 23297: `SUPPOSE YOU THROW A BASEBALL STRAIGHT UP AT A VELOCITY OF 64 FEET PER SECOND.
A FUNCTION
CAN BE CREATED BY EXPRESSING DISTANCE ABOVE GROUND, S, AS A FUNCTION OF TIME, T.
THIS FUNCTION IS: S=-16T^2+^V0^T+^S0
16 REPRESENTS 1/2G, THE GRAVITATIONAL PULL DUE TO GRAVITY (MEASURED IN FEET PER
SECOND^2).
^V0 IS THE INITIAL VELOCITY (HOW HARD DO YOU THROW THE OBJECT, MEASURED IN FEET
PER SECOND).
^S0 IS THE INITIAL DISTANCE ABOVE GROUND (IN FEET). IF YOU ARE STANDING ON THE
GROUND, THEN ^S0=0.

How long will it take to hit the ground?
What is the maximum height of the ball? Waht time was the maximum height attained?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The equation is:
s%28t%29+=+-16t%5E2+%2B+64t To find the time (t) at which the ball returns to the ground (s=0), set the above equation equalto 0 and solve for t.
0+=+-16t%5E2%2B64t Factor out a t.
0+=+t%28-16t%2B64%29 Apply the zero product principle.
t+=+0 and/or -64t%2B64+=+0
The solution t=0 would be the situation at the start of this operation. So we are left with:
-16t%2B64+=+0 Add 64t to both sides of the equation.
64+=+16t Divide both sides by 16.
t+=+4
The ball returns to the ground at t = 4 seconds.
To find the maximum height reached by the ball you first will find the time, t, at which the function s(t) is a maximum. Recall that the quadratic function represents a parabola and the maximum (or minimum) point on a parabola occurs at its vertex. The vertex, in this case, will be a maximum value because the parabola opens downward. How do you know this?...the coefficient of t^2 is negative.
The x-coordinate (or, in this case, the t-coordinate) of the vertex is given by:t+=+-b%2F2a where a=-16 and b=64. This is taken from the standard form for the quadratic equation: at%5E2%2Bbt%2Bc=0
The t-coordinate of the vertex is:
t+=+%28-64%29%2F2%28-16%29
t+=+%28-64%29%2F%28-32%29
t+=+2 The maximum height occurs at time t=2 seconds. Find the maximum height by substituting this value of t into the original function: s%28t%29+=+-16t%5E2%2B64t
s%282%29+=+-16%282%29%5E2+%2B+64%282%29
s%282%29+=+-64%2B128
s%282%29+=+64
The maximum height attained by the ball is 64 feet at 2 seconds.