A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t2 + 8 underroot t, where t is the time in seconds and s(t) is the distance in feet from the top of the building. A) Write a formula for the average velocity of the ball for t near 4. B) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds. C) What is your estimate for the instantaneous velocity of the ball at t = 4.

Explanation:

s(t) = -16t2 + 8√t

A )

Average velocity

s(t) / t = (-16t2 + 8√t)/t

A(t)= -16t +  8 / √t

average velocity of the ball for t near 4.

A(t) =  -16t +  8 / √t

          Lt t⇒4

B )

Distance covered in 4 s

-16t2 + 8√t

= - 16 x 16 + 8 x 2

= - 240

Distance covered in 5 s

= - 16 x 25 + 8 √5

= -400 + 17.88

= -382.12

distance covered in duration from 4 to 5 sec

= -142.12

velocity = - 142.12 / 1 = - 142.12 m /s

Distance covered in 4.5 s

= -16 x 4.5² + 8√4.5

= -324 + 16.97

= -307

distance covered during 4 to 4.5

= 67

velocity during 4 to 4.5

= -67 / .5

- 134 m /s

distance covered in 4.05 s

-16 x 4.05² + 8√4.05

-262.44 + 16.1

-246.34

distance covered during 4 to 4.05

= -6.34

velocity during 4 to 4.05

= -6.34 / .05

- 126.8 m /s

C )

Instantaneous velocity at t = 4

= - 120 m /s

momentum of soft ball three times less than baskey ball