Plz Help So I Can Finish?

Question

Suppose a baseball is shot up from the ground straight up with an initial velocity of 64 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 = -16t2 + v0t + s0
• 16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
• 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.
a) What is the function that describes this problem?

b) The ball will be how high above the ground after 1 second?

c) How long will it take to hit the ground?


d) What is the maximum height of the ball? What time will the maximum height be attained?

Answer 1

a) The function is correct, with s0=0 since it started at the ground, and v0=64 upward, which is the positive direction according to your sign convention

b)
s(1)=-16+64
=48 feet

c) to find the time that it will return to the ground,
set s(t)=0, since this is a quadratic, it will have two roots, take-off, t=0, and landing,
0=-16*t^2+64*t
64/16=t
=4 seconds

Maximum height may be found one of two ways.
Since the equation is symmetric about the apogee, the time to apogee will be 2 s
and the height will be
s(2)=-16*2+64*2
=128-32
=96 feet

another way to find (and check the answer) is to look at where
v(t)=0
v(t) is the first derivative of s(t)
v(t)=-32*t+64
set equal to 0 and
t=2 seconds

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