If you are standing near the edge of the top of a 200 feet building and throw a ball vertically upward it will be modeled by the function s(t)=-16t^2+64t+200 where s(t) is the ball's height above ground in feet and t is seconds after the ball was thrown.

Question

A) When does the ball reach its maximum height and what is the maximum height?
T= -b/2a
T= -64/-2(-16)
T= 2 seconds
How do I find the maximum height?
B) When does the ball hit the ground? (Round to the nearest tenth of a second)
Can you please help me with this question?
C) Find s(0) and describe what it represents?
Can you please help me with this question?
D) How would I graph the quadratic function
Please help me solving each parts step by step!!!!

Answer 1

A. Plug your calculated value of T(2 s.)
into the given Eq and solve for S, the ht.

B. Tr = 2 s. = Rise time(part A).

h = 0.5g*Tf^2 = 264 Ft.
g = 32Ft/s^2.
Tf = Fall time.
Tf = ?

Tr + Tf = Time to reach gnd.

C. S(o) = ht. when t = 0. In the given Eq, replace t with zero and solve for S(o).

D. Using the given Eq, calculate the ht.
for each value of T. Use the data for graphing.

(T, h).
(0.0,200).
(0.5, ).
(1.0,248).
(1.5, ).
V(2.0,264).
(2.5, ).
(3.0,248).
(3.5, ).
(4.0,200).