A baseball is tossed straight up reaches a height of s(t) = 8t 2 + 40t + 5 I hav

Question

A baseball is tossed straight up reaches a height of s(t) = 8t 2 + 40t + 5

I have found the functions of velocity and accel.

v(t)=-16t+40

a(t) or v'(t)=-16*

For part b I input the velocity at 0, and got 40ft/s

also put acceleration at 0 and got -16 ft/s^2, I am not sure this is correct.

I am confused about part c and do not know how to calculate that

For part d) I got time at 2.5 seconds and height at 57.5 feet by using t = -b/2a and ax^2+bx+c = 0

(2) A baseball is tossed straight up reaches a height of s(t) =-8t2 + 40t + 5, where s is feet above ground and t is seconds after leaving the players hand (a) Find the velocity and acceleration functions, (b) Find the velocity and acceleration immediately after the ball leaves the player's hand. Include units. (c) At what time(s) will the baseball be above 54ft? W1 (d) At what time will the baseball reach its maximum height? What is this height?

Explanation / Answer

s(t) = 8t2 + 40t + 5

c. 8t2 + 40t + 5>54

   8t2-40t +49<0

2.15<t<2.85

d. s(t) =8t2+40t +5

s'(t)= -16t +40

s"(t)=-16<0, So s(t) is maximum

s '(t)=0

-16t +40=0

t=2.5 sec

s(2.5)=-8(2.5)2+40(2.5)+5 =55feet

Maximum height is at t=2.5sec and maximum height is 55 feet

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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