Can you help me solve this - since it's Christmas? :)?

Question

s(t) = -16t^2 + v0t + s0

We throw a rock into the air with initial velocity of 96 ft/sec, and initial position of 5 ft. Find how high the rock goes before coming back down. To answer this question, fill in the blanks in the following sentence, where the first blank gives this height, and the second blank gives how many seconds into the flight of the rock it reaches this maximum height.

The rock reaches its highest position of _______ feet above the ground after _______ seconds of flight


I'll take anything I can get. Thanks so much! And.... MERRY CHRISTMAS!

Answer 1

First write your equation with the initial velocity (v0 = 96) and the initial height (s0 = 5) plugged in:

s(t) = -16t² + 96t + 5

Now factor out -16 from the first 2 terms:
s(t) = -16(t² - 6t) + 5

What we want to do is "complete the square". Take the coefficient on the t term (-6), halve it (-3) and square it (9). Add this and subtract it inside the parentheses:

s(t) = -16(t² - 6t + 9 - 9) + 5

Take the -9 outside the parentheses (by multiplying by -16):

s(t) = -16(t² - 6t + 9) -16(-9) + 5
s(t) = -16(t² - 6t + 9) + 144 + 5

Simplify the last part:
s(t) = -16(t² - 6t + 9) + 149

The expression in the parentheses is equivalent to (t - 3)(t - 3) because we completed the square. Rewrite the parentheses as a perfect square:
s(t) = -16(t - 3)² + 149

Notice that if t = 3, then the first expression is zero. That means we subtract nothing. That also means that 149 will be the maximum value of the height. Now you have your answers:

The rock reaches its highest position of --> 149 ft.
above the ground after --> 3 seconds
of flight.

P.S. This is the way to solve it without needing calculus. If you are familiar with calculus, you could also take the derivative and set it to zero. That will tell you the time t = 3. Plugging that back into the first equation will tell you the maximum height is 149 ft.

Answer 2

s(t) = -16t^2 + v0 t + s0

where s(t) = position of stone at time t in sec

v0 = intial velocity = 96 ft/sec

s0 = intial position = 5ft

the highest position will be when ds/dt = 0

s'(t) = -32t + vo

-32t + v0 = 0

v0 = 32t

since v0 = 96

32t = 96

t = 96/32 = 3 sec

substitute t= 3 in s(t) to get maximum height

s(3) = -16(9) + 96(3) + 5

s(3) = -144 + 288 + 5 = 149 ft

Answer 3

this is a parabolic trajectory, okay?!
s(t) = -16t² + v0t + s0
v0 = 96 and s0 = 5
so you have
s(t) = -16t² + 96t + 5
take the derivative and set equal to zero
s'(t) = 0 = - 32t + 96, solve for t
t = 3 seconds of flight,
so s(3) = -16(3)² + 96(3) + 5 = -144 + 288 + 5 = 149 ft high
MC & HNY!

Answer 4

Are you throwing it straight up? If that is true then 3 seconds later it stops. at=v. 16*3^2= 144 ft. add 5 ft. Thrown upward at roughly 65 mph First derivative of distance is velocity. since that is a known (96) and you know the acceleration (32 ft/s^2).