s=-1/2 gt2+v0 t+s0, where s is the height of the objected at time t seconds, g i

Question

s=-1/2 gt2+v0 t+s0, where s is the height of the objected at time t seconds, g is 32 feet per second2 and v0 and s0 are the initial velocity and initial height, respectively. Two friends are in competition. Each stands on the roof of the apartment building where each one lives. At the same time, each friend stands at the edge of the roof and throws up a basketball as hard as possible. Joan's apartment building is 120 feet high, and she trows the basketball at a speed of 24 feet per second. Joe's apartment building is 105 feet high, and he trows the basketball at a speed of 30 feet per second. a)write the function that describes the height of Joan's basketball, in terms of time measured in seconds. What kind of function is the height function? How do you know? b)Display the graph of the function you found from a) above. c) what is the maximum height of the basketball, and when does it reach that height? show your work. d) when does the basketball hit the ground? show your work. e) write the function that describes the height of Joe's basketball, in terms of time measured in seconds. f) which basketball reaches the a greater height? show work. g) is there a time when the basketballs are at the same height? set up the equation and use algebra to show work. h) which basketball stays up in the air longer? show work.

Explanation / Answer

a(t)=-32
v(t)=-32t +c
v(0)=24=-32(0)+c, so c=24
v(t)=-32t+24

s(t)=-16t^2+24t+c
s(0) = 120 = -16(0)^2 +24(0) + c, so c=120
s(t)=-16t^2+24t+120

It's a quadratic,because the highest power on t is 2.

c. s'(t) =-32t+24
-32t +24=0 ==> t=3/4sec

s(3/4)=-16(3/4)^2+24(3/4)+120
= 129 feet

d. -16t^2+24t+120=0
quadratic formula - use the positive root t=(3+sqrt(129))/4

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.