Word Problem, please help?

Question

Suppose a baseball is shot up from the ground straight up with an initial velocity of 32 feet per second. A function can be created by expressing distance above the ground,s, as a function of time,t. This function is s= -16t^2+ vot +so. 16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second^2). vo is the intitial velocity (how hard do you throw the object, measured in feet per second^2). so is the initial distance above ground (in feet). If you are standing on the ground, then so=0. What is the function that describes this problem? The ball will be how high above the ground after 1 second? How long will it take to hit the ground? What is the maximum height of the ball? What time will the maximum hieght be attained?

Answer 1

Initial velocity = Vo = 32 ft/sec
Initial height = So = 0 ft (starting from the ground)

V at time t = Vt = Vo - 32t
(each second the baseball goes 32 ft/sec slower)

S at time t = Vot -16t^2
S at (1 second) = 32 ft - 16 ft = 16 ft.

S(maximum) occurs when the Velocity = zero (before it starts downward)
V = zero when Vo - 32t = 0
32 = 32t, t = 1 second

The maximum height is 16 feet, which happens at t = 1 second.

Answer 2

s = -16t^2+32t

when t= 1 second, s=16

It hits the ground in only 2 seconds

The maximum height is 16 feet, 1 second after launch

Answer 3

s=-16t^2 +Vot + so
ds/dt =-32t +Vo
Setting this =0 gives -32t = -Vo so t = Vo/32
t=Vo/32 is the time at which the height is maximum, so
s= 16 (Vo/32)^2 +Vo(V0/32)
s= 16(Vo/32)[Vo/32 +Vo] =(Vo/2)[(Vo/32)+Vo]

After 1 second, s = -16 + Vo feet

The function is a parabola which is concave upward

When the ball hits the ground -16t^2 +Vot=0
So t(-16t+ Vo) = 0
This gives t=0 and t = Vo/16